201. In a diagnostic X-ray experiment, an object is placed between an X-ray tube and a detector. The source initially emits a uniform beam with intensity \(I_0\). An operator increases the incident beam intensity to \(2I_0\), while the object's transmission fractions remain unchanged. Region P transmits \(20\%\) and Region Q transmits \(60\%\) of the incident radiation. The detector is linear, with its response proportional to transmitted intensity. Which statement correctly predicts the effect of the intensity change?
ⓐ. The transmitted intensities double, but their ratio changes from \(1:3\) to \(1:6\).
ⓑ. The transmitted intensities remain unchanged because transmission fractions are material properties.
ⓒ. The transmitted intensities double, their ratio remains \(1:3\), and the absolute difference between them doubles.
ⓓ. The transmitted intensities become equal because a stronger beam eliminates differential absorption.
Correct Answer: The transmitted intensities double, their ratio remains \(1:3\), and the absolute difference between them doubles.
Explanation: \(\textbf{Determine initial transmitted intensities:}\)
For a region with transmission fraction \(T\),
\[
I_{\text{transmitted}}=TI_{\text{incident}}
\]
Initially, the incident intensity is
\[
I_{\text{incident}}=I_0
\]
For Region P,
\[
I_{P1}=0.20I_0
\]
For Region Q,
\[
I_{Q1}=0.60I_0
\]
\(\textbf{Compare the original responses:}\)
The ratio is
\[
\frac{I_{P1}}{I_{Q1}}=\frac{0.20I_0}{0.60I_0}
\]
\[
\frac{I_{P1}}{I_{Q1}}=\frac13
\]
The original absolute difference is
\[
\Delta I_1=I_{Q1}-I_{P1}
\]
\[
\Delta I_1=0.60I_0-0.20I_0
\]
\[
\Delta I_1=0.40I_0
\]
\(\textbf{Apply the increased source intensity:}\)
The new incident intensity is
\[
I'_{\text{incident}}=2I_0
\]
The transmission fractions remain unchanged.
Therefore,
\[
I_{P2}=0.20(2I_0)
\]
\[
I_{P2}=0.40I_0
\]
Similarly,
\[
I_{Q2}=0.60(2I_0)
\]
\[
I_{Q2}=1.20I_0
\]
\(\textbf{Compare the new responses:}\)
Their ratio is
\[
\frac{I_{P2}}{I_{Q2}}=\frac{0.40I_0}{1.20I_0}
\]
\[
\frac{I_{P2}}{I_{Q2}}=\frac13
\]
The new absolute difference is
\[
\Delta I_2=1.20I_0-0.40I_0
\]
\[
\Delta I_2=0.80I_0
\]
Thus,
\[
\frac{\Delta I_2}{\Delta I_1}=\frac{0.80I_0}{0.40I_0}=2
\]
\(\textbf{Physical interpretation:}\)
Doubling the incident intensity increases both transmitted intensities by the same factor.
It does not change the specified fractions transmitted through the two regions.
Consequently, the relative transmitted-intensity ratio remains unchanged.
However, the absolute difference in detector responses becomes twice as large.
This distinguishes relative contrast from the absolute signal difference.
The stronger beam also changes the radiation exposure involved, so increasing intensity is not automatically desirable in a medical setting.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{I_{P2}}{I_{Q2}}=\frac13,\qquad\frac{\Delta I_2}{\Delta I_1}=2}
\]
202. A medical facility uses electromagnetic radiation for different purposes. One instrument produces X-rays by directing energetic electrons onto a metal target and records their transmission through a patient's body. Another arrangement uses a carefully controlled high-energy radiation beam in the treatment of a selected tumour. Consider the following statements.
Statement I: The first arrangement can generate diagnostic images by detecting different X-ray transmissions through different tissues.
Statement II: X-rays can have medical applications beyond diagnosis, including selected cancer treatments.
Statement III: The biological effects of X-rays guarantee that every X-ray exposure produces the same degree of tissue damage, regardless of the exposure conditions.
Statement IV: The usefulness of high-energy electromagnetic radiation in medicine does not eliminate the need to control unnecessary exposure.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because different tissues can absorb or transmit different fractions of incident X-ray radiation.
A detector placed beyond the patient records the transmitted radiation.
Spatial differences in the detected signal can provide information about internal body structures.
This is the basis of diagnostic X-ray imaging.
Statement II is also correct because controlled X-ray radiation can be used in certain forms of medical treatment.
Such treatment differs in purpose from diagnostic imaging.
In diagnosis, the aim is to obtain information about the patient's internal structures.
In treatment, controlled radiation can be used to produce an intended effect in selected tissue.
Statement III is incorrect because biological damage does not occur to exactly the same extent under every exposure condition.
The delivered radiation exposure, distribution within tissue and biological context are relevant.
It is therefore unjustified to claim that all exposures necessarily produce identical damage.
Statement IV is correct because high-energy electromagnetic radiation can affect healthy tissue as well as tissue targeted for treatment.
Medical benefit does not make unnecessary irradiation harmless.
Suitable control of radiation exposure remains important in diagnostic and therapeutic applications.
The two medical uses consequently illustrate different objectives involving the same broad class of electromagnetic radiation.
The scientifically valid combination is Statements I, II and IV.
203. An X-ray imaging laboratory uses the same constant incident X-ray intensity in two successive exposures. In Exposure P, the beam passes through a region transmitting \(25\%\) of the incident radiation and then reaches a photographic film for \(12\,\text{s}\). In Exposure Q, the beam passes through another region transmitting \(50\%\), followed by a protective window transmitting \(60\%\) of the radiation reaching it. Assume linear film response, unchanged beam intensity, and that equal accumulated transmitted X-ray energy per unit area produces equal film exposure. For how long must Exposure Q continue to produce the same film exposure as P?
ⓐ. \(5.0\,\text{s}\)
ⓑ. \(10\,\text{s}\)
ⓒ. \(14.4\,\text{s}\)
ⓓ. \(24\,\text{s}\)
Correct Answer: \(10\,\text{s}\)
Explanation: \(\textbf{Identify the required film-exposure condition:}\)
The photographic films receive X-rays transmitted through their respective arrangements.
For constant transmitted intensity, accumulated exposure is proportional to
\[
H=I_{\text{transmitted}}t
\]
The two films must receive equal exposure:
\[
H_P=H_Q
\]
Let the common incident intensity be \(I_0\).
\(\textbf{Determine exposure through Region P:}\)
The transmitted fraction is
\[
T_P=0.25
\]
Therefore,
\[
I_P=0.25I_0
\]
The exposure duration is
\[
t_P=12\,\text{s}
\]
Hence,
\[
H_P=(0.25I_0)(12)
\]
\[
H_P=3I_0\,\text{s}
\]
\(\textbf{Account for both attenuating elements in Q:}\)
The object region transmits
\[
T_{\text{object}}=0.50
\]
The protective window transmits
\[
T_{\text{window}}=0.60
\]
Because the window acts on radiation already transmitted through the object, the overall transmitted fraction is
\[
T_Q=T_{\text{object}}T_{\text{window}}
\]
\[
T_Q=(0.50)(0.60)
\]
\[
T_Q=0.30
\]
Consequently,
\[
I_Q=0.30I_0
\]
\(\textbf{Solve for the required exposure time:}\)
\[
I_Qt_Q=H_P
\]
Substituting,
\[
(0.30I_0)t_Q=3I_0\,\text{s}
\]
Canceling the common incident intensity,
\[
0.30t_Q=3\,\text{s}
\]
\[
t_Q=10\,\text{s}
\]
\(\textbf{Interpret the compensation:}\)
Although Region Q alone transmits twice the fraction transmitted by P, its additional window reduces the total fraction reaching the film.
The net transmission through Q is \(30\%\), compared with \(25\%\) through P.
Therefore, Q requires a somewhat shorter exposure time.
Neither object transmission nor window transmission can be ignored.
The calculation concerns the energy recorded by the film, not the biological safety of either arrangement.
Equal detector exposure does not necessarily imply equal absorption in the objects.
\(\textbf{Final answer:}\)
\[
\boxed{t_Q=10\,\text{s}}
\]
204. An X-ray tube contains an electron beam directed towards a metal target. The radiation emerging from the target passes through a separate evacuated region before reaching a detector. Two experiments are conducted. In Experiment P, a uniform magnetic field perpendicular to the electron velocity is introduced between the electron source and target. In Experiment Q, the same magnetic field is instead introduced only in the region between the target and X-ray detector. Neglect interactions with the apparatus and assume the target's X-ray output is otherwise unchanged in Q. Which prediction is correct?
ⓐ. P can alter the electron trajectory before impact, whereas Q does not bend the X-rays merely by applying the static magnetic field.
ⓑ. P cannot alter the electron trajectory because electrons and X-rays travel at the same speed.
ⓒ. Q necessarily bends the X-rays more strongly than P bends the electrons because X-rays have higher frequencies.
ⓓ. Both fields necessarily bend the X-ray beam because electromagnetic waves contain magnetic-field components.
Correct Answer: P can alter the electron trajectory before impact, whereas Q does not bend the X-rays merely by applying the static magnetic field.
Explanation: The experiment distinguishes the charged particles used to generate X-rays from the electromagnetic radiation subsequently emitted.
The incident electron beam consists of negatively charged particles.
A magnetic field can exert a magnetic force on a moving charged particle.
When the magnetic field is perpendicular to the electron velocity, the force can change the electron's direction of motion.
Thus, introducing the field before the target can alter where or how the electrons strike it.
This may affect the subsequent X-ray production arrangement.
The radiation generated when energetic electrons interact with the metal target is different from the incident electron beam.
X-rays are electromagnetic waves.
They do not consist of the original electrons continuing along the radiation path.
A freely propagating X-ray beam in vacuum is not bent by an ordinary static magnetic field merely because its frequency is high.
The wave's own magnetic-field component must not be confused with a material particle carrying electric charge.
Introducing the external field only between target and detector therefore does not produce the same charged-particle deflection mechanism.
The stipulated unchanged target output in Q isolates propagation from production.
The experiment consequently separates the magnetic response of an incident electron beam from the vacuum propagation of the emitted X-rays.
The correct prediction is that P can alter electron trajectories, whereas Q does not bend the X-rays through the stated static-field mechanism.
205. An excited nucleus emits gamma radiation, which travels through vacuum and enters a gas-filled radiation detector. The detector produces an electrical signal when the radiation interacts with the gas. Consider the following statements.
Statement I: The emitted gamma radiation is electromagnetic even though its source is a nucleus.
Statement II: The gamma ray must itself carry electric charge because the detector eventually produces an electric current.
Statement III: The detector response can arise from ionisation produced by interaction of gamma radiation with matter.
Statement IV: The emitted gamma ray requires nuclei to move continuously from source to detector to transport its energy.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements II and IV only
ⓒ. Statements I and III only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I and III only
Explanation: Statement I is correct because gamma rays are electromagnetic radiation.
They are characteristically associated with nuclear processes, including transitions of excited nuclei.
Their nuclear origin does not make them streams of nuclei.
Once emitted, gamma radiation can propagate through vacuum as an electromagnetic disturbance.
Statement II is incorrect because the gamma ray does not need to carry electric charge to produce an electrical detector response.
An incident electromagnetic wave can transfer energy to matter.
Gamma radiation interacting with a suitable gas can produce ionisation.
The resulting charged particles may then be collected by the detector's electrical arrangement.
Thus, the detector current is not evidence that the incident gamma ray itself carried charge.
Statement III is correct because ionisation-based detection is a suitable method for registering high-frequency electromagnetic radiation.
The detector converts an interaction of radiation with matter into a measurable signal.
Statement IV is incorrect because gamma radiation transports energy without requiring nuclei to travel from the source to the receiver.
The source nucleus and the emitted electromagnetic radiation are physically distinct.
Gamma rays also propagate through vacuum at the same speed as other electromagnetic waves.
The experiment therefore combines nuclear emission, electromagnetic propagation and matter-based detection.
Only Statements I and III correctly describe the process.
206. A gamma-ray detector registers radiation from a radioactive source against a steady background. Its readings are shown below.
| Condition | Detector count rate |
|---|
| Source absent | \(20\,\text{s}^{-1}\) |
| Source present, no absorber | \(220\,\text{s}^{-1}\) |
| Source present, absorber P | \(140\,\text{s}^{-1}\) |
| Source present, absorber Q | \(60\,\text{s}^{-1}\) |
Assume that background is unchanged and that the corrected source count rate is proportional to the transmitted gamma-ray intensity. What are the fractions of the source radiation transmitted through P and Q?
ⓐ. \(7/11,\;3/11\)
ⓑ. \(3/5,\;1/5\)
ⓒ. \(2/5,\;4/5\)
ⓓ. \(7/10,\;3/10\)
Correct Answer: \(3/5,\;1/5\)
Explanation: \(\textbf{Identify the background contribution:}\)
The detector records some counts even when the gamma source is absent.
The background count rate is
\[
R_b=20\,\text{s}^{-1}
\]
This contribution is present during every source measurement.
Therefore, raw detector readings must not be used directly as ratios of transmitted source intensity.
\(\textbf{Determine the unobstructed source response:}\)
With the source present but without an absorber,
\[
R_0=220\,\text{s}^{-1}
\]
Subtracting background,
\[
R_{\text{source},0}=R_0-R_b
\]
\[
R_{\text{source},0}=220-20
\]
\[
R_{\text{source},0}=200\,\text{s}^{-1}
\]
\(\textbf{Correct the measurement through P:}\)
The raw count rate is
\[
R_P=140\,\text{s}^{-1}
\]
The corrected source count rate is
\[
R_{\text{source},P}=140-20
\]
\[
R_{\text{source},P}=120\,\text{s}^{-1}
\]
The transmitted fraction is
\[
T_P=\frac{R_{\text{source},P}}{R_{\text{source},0}}
\]
\[
T_P=\frac{120}{200}
\]
\[
T_P=\frac35
\]
\(\textbf{Correct the measurement through Q:}\)
The raw count rate is
\[
R_Q=60\,\text{s}^{-1}
\]
Subtracting background,
\[
R_{\text{source},Q}=60-20
\]
\[
R_{\text{source},Q}=40\,\text{s}^{-1}
\]
Therefore,
\[
T_Q=\frac{40}{200}
\]
\[
T_Q=\frac15
\]
\(\textbf{Interpret the attenuation:}\)
Absorber P transmits \(60\%\) of the source-related detector signal.
Absorber Q transmits only \(20\%\).
Q therefore produces greater attenuation under the specified conditions.
The raw readings include contributions unrelated to the source and cannot directly represent transmission percentages.
This is especially important when the transmitted source signal is small compared with background.
The calculation uses a linear detector model and does not require any detailed attenuation law.
\(\textbf{Final answer:}\)
\[
\boxed{(T_P,T_Q)=\left(\frac35,\frac15\right)}
\]
207. An excited nucleus emits monochromatic gamma radiation with vacuum wavelength \(3.0\times10^{-12}\,\text{m}\). A second source emits electromagnetic radiation at frequency \(2.0\times10^{19}\,\text{Hz}\). Assume that the second radiation travels through vacuum and that its production mechanism is not specified. Taking \(c=3.0\times10^8\,\text{m s}^{-1}\), which conclusion is correct?
ⓐ. The gamma radiation has frequency \(1.0\times10^{20}\,\text{Hz}\), and the two sources must have identical nuclear production mechanisms.
ⓑ. The gamma-ray frequency is \(1.0\times10^{20}\,\text{Hz}\), five times the second frequency; the second source's mechanism remains unspecified.
ⓒ. The gamma radiation has frequency \(1.0\times10^{20}\,\text{Hz}\), one-fifth that of the second radiation, and must travel more slowly.
ⓓ. The gamma radiation has frequency \(1.0\times10^{19}\,\text{Hz}\), and the second radiation must be ultraviolet.
Correct Answer: The gamma-ray frequency is \(1.0\times10^{20}\,\text{Hz}\), five times the second frequency; the second source's mechanism remains unspecified.
Explanation: \(\textbf{Calculate the frequency of the nuclear radiation:}\)
The gamma-ray wavelength is
\[
\lambda_\gamma=3.0\times10^{-12}\,\text{m}
\]
In vacuum,
\[
c=f_\gamma\lambda_\gamma
\]
Therefore,
\[
f_\gamma=\frac{c}{\lambda_\gamma}
\]
Substituting,
\[
f_\gamma=\frac{3.0\times10^8}{3.0\times10^{-12}}
\]
\[
f_\gamma=1.0\times10^{20}\,\text{Hz}
\]
\(\textbf{Compare the two frequencies:}\)
The second source has frequency
\[
f_2=2.0\times10^{19}\,\text{Hz}
\]
Thus,
\[
\frac{f_\gamma}{f_2}=\frac{1.0\times10^{20}}{2.0\times10^{19}}
\]
\[
\frac{f_\gamma}{f_2}=5
\]
The gamma radiation has five times the frequency of the second beam.
\(\textbf{Examine what the frequency does not establish:}\)
The first source is explicitly described as an excited nucleus.
This is a characteristic source of gamma radiation.
However, the second source's production mechanism is not specified.
Its high frequency alone does not establish that its radiation was produced by a nuclear transition.
X-ray and gamma-ray wavelength regions may overlap.
A high-frequency electromagnetic wave can therefore require source information for a precise source-based classification.
\(\textbf{Check propagation behaviour:}\)
Both waves travel through vacuum.
Their propagation speeds are consequently equal:
\[
v_\gamma=v_2=c
\]
Their frequency difference is associated with different wavelengths, not different vacuum speeds.
\(\textbf{Final answer:}\)
\[
\boxed{f_\gamma=1.0\times10^{20}\,\text{Hz},\qquad\frac{f_\gamma}{f_2}=5}
\]
208. A laboratory directs two beams through an evacuated region containing a uniform electric field perpendicular to their initial directions of travel. Beam P is gamma radiation emitted by excited nuclei. Beam Q consists of freely moving charged particles produced by a separate source. After leaving the field region, both beams enter suitable detectors. Ignore gravitational effects and interactions with residual gas. Which observation is physically consistent with the stated conditions?
ⓐ. P must bend more strongly because its radiation frequency is extremely high, while Q remains straight because its particles are material.
ⓑ. Both beams must bend equally because both can subsequently produce ionisation in detector matter.
ⓒ. P cannot propagate through the evacuated region because it has no charge carriers.
ⓓ. Q can be deflected by the applied electric field, while P continues without electrostatic bending and can still produce ionisation upon detection.
Correct Answer: Q can be deflected by the applied electric field, while P continues without electrostatic bending and can still produce ionisation upon detection.
Explanation: The experiment separates electromagnetic radiation from a stream of electrically charged material particles.
Beam P consists of gamma rays emitted by excited nuclei.
Gamma rays are electromagnetic radiation.
They are not nuclei or charged particles travelling away from the source.
Consequently, an ordinary static electric field does not produce the charged-particle deflection mechanism for Beam P.
The gamma radiation can therefore continue along its original path through the evacuated region under the stated ideal conditions.
Beam Q contains freely moving charged particles.
An electric field exerts a force on an electrically charged particle.
If the force has a component perpendicular to its velocity, its trajectory can change.
Thus, Q can be deflected by the applied field.
Both beams may subsequently produce electrical signals in suitable detectors.
However, similar detector responses do not require the incident beams to possess the same electrical charge.
Gamma radiation can produce ionisation through interaction with the detector material.
Beam Q may produce ionisation through interactions involving its charged particles.
The important distinction is between charge carried by the incident beam and charged particles generated during detection.
Therefore, electrostatic deflection can distinguish the charged-particle beam from gamma radiation in the specified arrangement.
209. Two gamma-ray absorbers P and Q are placed in succession in the path of a beam carrying \(100\,\text{J}\) of incident electromagnetic energy. Absorber P transmits \(80\%\) of the energy incident upon it, and Absorber Q transmits \(50\%\) of the energy incident upon it. Assume that energy not transmitted is absorbed and that the stated transmission fractions are unchanged when the absorber order is reversed. Which statement correctly compares the arrangements P → Q and Q → P?
ⓐ. Both arrangements transmit \(40\,\text{J}\), but the energy absorbed by P differs between the two arrangements.
ⓑ. P → Q transmits \(40\,\text{J}\), while Q → P transmits \(50\,\text{J}\).
ⓒ. Both arrangements transmit \(40\,\text{J}\), and each individual absorber necessarily absorbs the same energy in both orders.
ⓓ. P → Q transmits \(80\,\text{J}\), while Q → P transmits \(50\,\text{J}\).
Correct Answer: Both arrangements transmit \(40\,\text{J}\), but the energy absorbed by P differs between the two arrangements.
Explanation: \(\textbf{Follow the beam through P first:}\)
The initial gamma-ray energy is
\[
U_0=100\,\text{J}
\]
Absorber P transmits a fraction
\[
T_P=0.80
\]
Therefore, the energy leaving P is
\[
U_{P,\text{out}}=0.80(100)
\]
\[
U_{P,\text{out}}=80\,\text{J}
\]
P absorbs the remaining energy:
\[
U_{P,\text{abs}}=100-80
\]
\[
U_{P,\text{abs}}=20\,\text{J}
\]
\(\textbf{Apply Q to the reduced beam:}\)
Q transmits \(50\%\) of the energy incident upon it.
Therefore,
\[
U_{PQ,\text{out}}=0.50(80)
\]
\[
U_{PQ,\text{out}}=40\,\text{J}
\]
Q absorbs
\[
U_{Q,\text{abs}}=80-40
\]
\[
U_{Q,\text{abs}}=40\,\text{J}
\]
\(\textbf{Reverse the absorber order:}\)
In the arrangement Q → P, Q receives the original \(100\,\text{J}\).
It transmits
\[
U_{Q,\text{out}}=0.50(100)
\]
\[
U_{Q,\text{out}}=50\,\text{J}
\]
Q therefore absorbs
\[
100-50=50\,\text{J}
\]
P now receives only \(50\,\text{J}\).
It transmits
\[
U_{QP,\text{out}}=0.80(50)
\]
\[
U_{QP,\text{out}}=40\,\text{J}
\]
P absorbs
\[
U'_{P,\text{abs}}=50-40
\]
\[
U'_{P,\text{abs}}=10\,\text{J}
\]
\(\textbf{Identify the invariant and changed quantities:}\)
The total transmitted energy is \(40\,\text{J}\) in either order because
\[
(0.80)(0.50)=(0.50)(0.80)
\]
However, P absorbs \(20\,\text{J}\) when placed first and \(10\,\text{J}\) when placed second.
Likewise, Q absorbs \(40\,\text{J}\) when second but \(50\,\text{J}\) when first.
Thus, unchanged overall transmission does not imply unchanged energy absorption in each individual absorber.
The result depends on applying each transmission fraction to the radiation actually reaching that absorber.
\(\textbf{Final answer:}\)
\[
\boxed{U_{PQ,\text{out}}=U_{QP,\text{out}}=40\,\text{J},\quad U_{P,\text{abs}}:U'_{P,\text{abs}}=2:1}
\]
210. A medical team considers the use of controlled gamma radiation to treat a malignant region. One proposal directs the radiation towards the intended target while limiting unnecessary irradiation of surrounding healthy tissue. Another proposal claims that no protection of surrounding tissue is needed because gamma rays can affect only cancerous cells. Which assessment is most appropriate?
ⓐ. The second proposal is correct because gamma rays become harmless immediately after entering healthy tissue.
ⓑ. The first proposal is appropriate because gamma exposure can affect healthy as well as malignant tissue.
ⓒ. Both proposals are equivalent because gamma rays cannot transfer energy to biological matter.
ⓓ. The first proposal is incorrect because gamma radiation can be used only for diagnosis, never for treatment.
Correct Answer: The first proposal is appropriate because gamma exposure can affect healthy as well as malignant tissue.
Explanation: Gamma radiation is high-frequency electromagnetic radiation.
It can be emitted in nuclear processes, including transitions of excited nuclei.
Gamma rays can transfer energy to matter through interactions with it.
Living tissues are not exempt from these interactions.
Controlled gamma radiation can be used in selected cancer-treatment applications.
The intended effect involves damaging malignant cells under carefully chosen conditions.
However, gamma radiation does not identify cancerous cells solely because they are malignant.
Healthy tissue exposed to the radiation can also experience biological effects.
Therefore, therapeutic usefulness does not imply complete selectivity or harmlessness outside the target.
The first proposal recognises the need to direct and control the radiation exposure.
The second proposal incorrectly assumes that all healthy tissues are unaffected.
Gamma radiation is neither universally harmless nor unsuitable for every medical treatment.
Its usefulness and risks depend on the purpose and circumstances of exposure.
The problem does not provide the information needed to determine a treatment schedule or biological response quantitatively.
It does support the principle that intended therapeutic irradiation must be distinguished from unnecessary exposure.
Thus, the first proposal is scientifically defensible.
211. An ionisation-based detector is used to register gamma radiation emitted by a radioactive source. A researcher makes two observations:
In Experiment P, the gamma source remains in a fixed position while the electric field used to collect ions inside the detector is reversed. The sign of the recorded electrical current reverses. In Experiment Q, the detector collection field is restored to its original direction, but a radiation absorber is placed between the gamma source and detector. The magnitude of the recorded source-related current decreases.
Which interpretation is most consistent with both observations?
ⓐ. Reversing the detector field reverses the charge carried by each incident gamma ray, while the absorber reduces gamma-ray speed in vacuum.
ⓑ. The gamma rays are moving ions, so reversing the collection field necessarily reverses their direction before they reach the detector.
ⓒ. The source emits visible light rather than gamma rays because a current is measured.
ⓓ. The detector current depends on collection of charged particles produced within the detector; reducing the radiation reaching it can reduce ionisation and current.
Correct Answer: The detector current depends on collection of charged particles produced within the detector; reducing the radiation reaching it can reduce ionisation and current.
Explanation: Gamma radiation is electromagnetic radiation emitted by suitable nuclear processes.
The incident gamma rays are not electrically charged material particles.
Nevertheless, gamma rays can interact with matter and cause ionisation.
A suitable gas-filled detector can collect the charged particles produced by such interactions.
Its electrical signal therefore arises through the detector's material response.
The external collection field influences the motion of the ions and electrons inside the detector.
Changing the direction of this collection field can change the direction assigned to the measured collection current.
Such a change does not imply that the incoming gamma rays have acquired or reversed electric charge.
In Experiment Q, an absorber is introduced before the radiation reaches the detector.
The absorber removes a fraction of the incident gamma radiation from the transmitted beam.
Under otherwise unchanged conditions, less radiation reaching the detector can produce fewer ionisation events.
This can reduce the source-related detector signal.
The two experiments consequently alter different parts of the measurement process.
Experiment P changes how detector-generated charges are collected.
Experiment Q changes how much electromagnetic radiation reaches the detector.
Neither observation requires gamma rays to be charged particles or to have a different vacuum propagation speed.
The correct explanation distinguishes radiation transport, ionisation and electrical charge collection.
212. A radioactive source emits gamma radiation towards two identical detectors. Detector P is \(2.0\,\text{m}\) from the source, while Detector Q is \(4.0\,\text{m}\) away. Assume the source emits electromagnetic energy uniformly in all directions, that the radiation spreads over spherical surfaces, and that the detectors intercept equal small areas perpendicular to their respective incident beams. A protective screen transmitting \(50\%\) of incident gamma energy is placed immediately before Q only. Neglect other absorption. What is the ratio of gamma energies received by P and Q during equal observation times?
ⓐ. \(2:1\)
ⓑ. \(4:1\)
ⓒ. \(8:1\)
ⓓ. \(16:1\)
Correct Answer: \(8:1\)
Explanation: \(\textbf{Identify the geometrical spreading condition:}\)
The source is specified to emit uniformly in all directions.
At distance \(r\), its radiation is distributed over a spherical surface of area
\[
A_s=4\pi r^2
\]
For equal observation times and equal small collecting areas, the received energy is inversely proportional to this spherical area.
Therefore,
\[
U\propto\frac1{r^2}
\]
before introducing an absorber.
\(\textbf{Compare the unshielded received energies:}\)
The detector distances are
\[
r_P=2.0\,\text{m}
\]
and
\[
r_Q=4.0\,\text{m}
\]
Without the protective screen,
\[
\frac{U_P}{U_{Q,0}}=\frac{r_Q^2}{r_P^2}
\]
Substituting,
\[
\frac{U_P}{U_{Q,0}}=\frac{(4.0)^2}{(2.0)^2}
\]
\[
\frac{U_P}{U_{Q,0}}=\frac{16}{4}
\]
\[
\frac{U_P}{U_{Q,0}}=4
\]
Thus, P receives four times the energy that Q would receive without screening.
\(\textbf{Apply the additional screen at Q:}\)
The screen transmits
\[
T=0.50
\]
of the energy incident upon it.
Therefore,
\[
U_Q=0.50U_{Q,0}
\]
\(\textbf{Determine the final ratio:}\)
\[
\frac{U_P}{U_Q}=\frac{U_P}{0.50U_{Q,0}}
\]
\[
\frac{U_P}{U_Q}=\frac4{0.50}
\]
\[
\frac{U_P}{U_Q}=8
\]
\(\textbf{Physical check:}\)
Doubling the distance reduces the intercepted energy by a factor of four for the specified isotropic source.
The screen then reduces Q's energy by another factor of two.
The factors combine to give an eightfold difference.
This does not mean that gamma-ray propagation speed decreases with distance.
The radiation spreads over a larger area, so a fixed detector intercepts a smaller share of the emitted energy.
The screen introduces a separate reduction through material interaction.
Geometrical spreading and absorption must therefore be treated as independent causes.
\(\textbf{Final answer:}\)
\[
\boxed{U_P:U_Q=8:1}
\]
213. A physics student makes the following claims about gamma-ray penetration:
Statement I: Gamma radiation can pass through some materials while being partly absorbed by others.
Statement II: Because gamma rays can be highly penetrating, no material can ever absorb them.
Statement III: A reduced gamma-ray detector reading behind a suitable absorber can result from removal of radiation from the transmitted beam.
Statement IV: If gamma radiation is absorbed by matter, it must have travelled more slowly than visible light through vacuum before entering the material.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I and III only
ⓒ. Statements II and IV only
ⓓ. Statements I, III and IV only
Correct Answer: Statements I and III only
Explanation: Statement I is correct because the interaction of gamma radiation with matter depends on the material and the radiation conditions.
Gamma rays are often described as penetrating electromagnetic radiation.
However, penetration does not mean that every gamma ray necessarily passes through every material without interaction.
Some fraction of incident gamma radiation can be absorbed or otherwise removed from a transmitted beam.
Statement II is incorrect because high penetration and complete immunity to absorption are different ideas.
A sufficiently suitable absorbing arrangement can reduce the gamma radiation reaching a detector.
Statement III is correct because a detector placed behind an absorber registers radiation that emerges from the material.
If the absorber reduces transmitted gamma intensity, the detector can register a smaller source-related signal.
This observation is consistent with radiation–matter interaction.
It does not require the original gamma rays to change into charged particles before entering the absorber.
Statement IV is incorrect because gamma radiation and visible light share the same propagation speed in vacuum.
Their different interactions with matter do not establish different vacuum speeds.
A gamma beam can travel through vacuum at \(c\) and subsequently lose part of its transmitted energy through absorption.
The relevant distinction is between free propagation and interaction with a material.
Therefore, Statements I and III alone are correct.
214. An oscillating radio transmitter produces a sinusoidal electromagnetic wave at frequency \(10^8\,\text{Hz}\). A separate radioactive source produces gamma-ray emissions from excited nuclei at an average rate of \(10^4\) emissions per second. A student argues that the gamma radiation must have frequency \(10^4\,\text{Hz}\), since that is the number of emission events per second. Which conclusion is correct?
ⓐ. The student is correct because the rate of nuclear emission events always equals the frequency of each emitted electromagnetic wave.
ⓑ. The student is correct because gamma rays have lower frequencies than radio waves whenever their sources are radioactive.
ⓒ. The student is incorrect only because gamma radiation travels faster than radio waves in vacuum.
ⓓ. The student is incorrect because the gamma emission-event rate and the oscillation frequency of the emitted electromagnetic radiation are distinct quantities.
Correct Answer: The student is incorrect because the gamma emission-event rate and the oscillation frequency of the emitted electromagnetic radiation are distinct quantities.
Explanation: The radio transmitter is driven by a harmonic electrical oscillation.
Its radiation frequency is related to the frequency of the oscillating charge motion.
The specified radio-wave frequency is therefore \(10^8\,\text{Hz}\).
The gamma-ray source operates through a different microscopic process.
Excited or radioactive nuclei can emit gamma radiation through nuclear transitions.
The average emission rate counts how frequently such emission events occur.
It does not specify the oscillation frequency associated with the electromagnetic radiation emitted in each event.
The number of separate emissions per second and the frequency of the emitted electromagnetic wave describe different physical properties.
For example, changing the number of nuclei contributing emissions can change the number of gamma-ray events detected without necessarily changing the characteristic emitted wavelength.
Gamma radiation characteristically occupies a very-high-frequency region of the electromagnetic spectrum.
Its wave frequency is not obtained simply by equating it with the average number of nuclear emissions each second.
A detector's count rate similarly measures detected events, not individual electromagnetic oscillations within each detected radiation event.
The radio and gamma sources therefore illustrate why a source's repetition rate must be interpreted according to its actual physical meaning.
Both kinds of radiation nevertheless propagate at the same speed through vacuum.
The student's inference confuses emission-event statistics with electromagnetic-wave frequency.
215. Four devices are proposed for a laboratory demonstration. Each device must use radiation within the frequency range accepted by its detector. The operating information is given below.
| Device | Source frequency | Required radiation and detector |
|---|
| P | \(5.0\times10^9\,\text{Hz}\) | Microwave radar receiver designed for GHz radiation |
| Q | \(5.0\times10^{14}\,\text{Hz}\) | Infrared detector specified to respond only to wavelengths longer than \(1\,\mu\text{m}\) |
| R | \(3.0\times10^{13}\,\text{Hz}\) | Infrared detector sensitive near \(10\,\mu\text{m}\) |
| S | \(6.0\times10^{14}\,\text{Hz}\) | Radio receiver sensitive only to frequencies near \(100\,\text{MHz}\) |
Assuming vacuum propagation with \(c=3.0\times10^8\,\text{m s}^{-1}\), which devices have compatible source and detector specifications?
ⓐ. P and R only
ⓑ. P, Q and R only
ⓒ. Q and S only
ⓓ. R and S only
Correct Answer: P and R only
Explanation: \(\textbf{Evaluate Device P:}\)
The source frequency is
\[
f_P=5.0\times10^9\,\text{Hz}
\]
This is a gigahertz-frequency electromagnetic wave.
Its vacuum wavelength is
\[
\lambda_P=\frac{c}{f_P}
\]
\[
\lambda_P=\frac{3.0\times10^8}{5.0\times10^9}
\]
\[
\lambda_P=0.060\,\text{m}
\]
\[
\lambda_P=6.0\,\text{cm}
\]
This wavelength lies in the microwave region, consistent with the specified radar receiver.
\(\textbf{Evaluate Device Q:}\)
\[
f_Q=5.0\times10^{14}\,\text{Hz}
\]
Therefore,
\[
\lambda_Q=\frac{3.0\times10^8}{5.0\times10^{14}}
\]
\[
\lambda_Q=6.0\times10^{-7}\,\text{m}
\]
\[
\lambda_Q=0.60\,\mu\text{m}
\]
This is a visible-light wavelength.
The detector is specified to respond only to wavelengths longer than \(1\,\mu\text{m}\).
Consequently, Q is incompatible with that detector specification.
\(\textbf{Evaluate Device R:}\)
\[
f_R=3.0\times10^{13}\,\text{Hz}
\]
Its wavelength is
\[
\lambda_R=\frac{3.0\times10^8}{3.0\times10^{13}}
\]
\[
\lambda_R=1.0\times10^{-5}\,\text{m}
\]
\[
\lambda_R=10\,\mu\text{m}
\]
This lies in the infrared region and matches the stated detector sensitivity.
Therefore, R is compatible.
\(\textbf{Evaluate Device S:}\)
The source frequency is
\[
f_S=6.0\times10^{14}\,\text{Hz}
\]
The specified receiver operates near
\[
100\,\text{MHz}=1.0\times10^8\,\text{Hz}
\]
The source frequency is therefore far outside the receiver's operating region.
A visible-frequency electromagnetic wave does not become a radio-frequency signal simply because a radio receiver is placed in its path.
\(\textbf{Physical interpretation:}\)
The successful arrangements require compatibility between the source radiation and the detector's spectral sensitivity.
Vacuum propagation speed alone cannot guarantee detector compatibility.
Both frequency and wavelength can be used to identify the relevant spectral region.
The result requires converting the source data into the quantities used to specify the detectors.
Devices P and R alone satisfy the stated conditions.
216. A laboratory must select a radiation source for three tasks:
P. Detecting a distant aircraft using a returned directional electromagnetic signal.
Q. Recording internal bone–soft-tissue contrast using transmission through the body.
R. Receiving a coded signal from a nearby hand-held controller using a detector sensitive to radiation around \(900\,\text{nm}\).
The available sources are:
Source 1: A \(10\,\text{GHz}\) microwave source.
Source 2: An electron-target X-ray tube.
Source 3: An infrared-emitting LED operating near \(900\,\text{nm}\).
Source 4: A visible lamp centred around \(550\,\text{nm}\).
Which source assignment is compatible with all three tasks?
ⓐ. P–2, Q–1, R–3
ⓑ. P–1, Q–3, R–4
ⓒ. P–1, Q–2, R–3
ⓓ. P–4, Q–2, R–1
Correct Answer: P–1, Q–2, R–3
Explanation: Task P requires a directional electromagnetic signal suitable for detecting a distant object through its returned radiation.
Microwaves are commonly used in radar because their relatively short wavelengths support directional transmission and reflected-signal detection.
The \(10\,\text{GHz}\) source therefore matches P.
Task Q requires radiation capable of producing a transmission-based image showing differences between bone-like and soft-tissue-like regions.
X-rays can penetrate body tissues to different extents.
The resulting differences in transmitted intensity can be recorded by a suitable detector.
An electron-target X-ray tube is consequently appropriate for Q.
Task R requires a coded signal to be received by a detector sensitive near \(900\,\text{nm}\).
That wavelength lies beyond the red end of the visible spectrum, within the infrared region.
An infrared-emitting LED operating near the detector's sensitive wavelength is therefore appropriate for R.
The visible lamp at \(550\,\text{nm}\) does not satisfy the stated remote-control detector condition.
The microwave source is not interchangeable with the X-ray tube merely because both emit electromagnetic radiation.
Their characteristic wavelengths and interactions with matter differ.
Likewise, the infrared LED is appropriate for signalling to the specified receiver but not for replacing X-rays in transmission imaging.
The source must be selected according to both its spectral properties and the intended physical interaction.
The correct assignment is P–1, Q–2 and R–3.
217. A manufacturer claims that its detector can distinguish every type of electromagnetic radiation solely by measuring vacuum propagation speed. The detector is tested with four monochromatic sources: a radio-frequency aerial, an infrared lamp, an ultraviolet lamp and a nuclear gamma-ray emitter. All four beams travel the same distance through vacuum. Which modification would most directly make the proposed classification physically meaningful?
ⓐ. Measure only the arrival times more accurately, because each spectrum region has a unique vacuum speed.
ⓑ. Determine frequency or vacuum wavelength and, where source-based labels overlap, consider how the radiation was produced.
ⓒ. Measure the beam's visible brightness, because every electromagnetic wavelength has a unique brightness visible to the human eye.
ⓓ. Measure whether each wave contains an electric field, because gamma rays possess only magnetic fields.
Correct Answer: Determine frequency or vacuum wavelength and, where source-based labels overlap, consider how the radiation was produced.
Explanation: All four tested beams are electromagnetic radiation.
In vacuum, electromagnetic waves propagate at the same speed \(c\), regardless of their frequency.
Therefore, measuring propagation speed alone cannot distinguish radio waves from infrared, ultraviolet or gamma radiation.
Even a highly accurate travel-time measurement would yield the same speed for all four beams over equal vacuum distances.
The electromagnetic spectrum is primarily ordered by frequency and wavelength.
These quantities satisfy the relation \(c=f\lambda\).
A frequency measurement can therefore identify the corresponding vacuum wavelength.
The measured value can then be compared with the characteristic regions of the electromagnetic spectrum.
Radio waves generally occupy much lower frequencies than infrared, ultraviolet or gamma radiation.
However, the boundaries between neighbouring spectral regions are not all perfectly sharp.
In particular, X-ray and gamma-ray wavelength ranges may overlap.
In an overlapping region, knowledge of the production mechanism can help distinguish the conventional source-based labels.
For example, energetic electrons striking a metal target characteristically produce X-rays, whereas excited nuclei characteristically emit gamma radiation.
Visible brightness is not a universal classification tool because much electromagnetic radiation is invisible to the human eye.
Nor can the presence of electric and magnetic fields distinguish spectrum regions, since all electromagnetic waves possess both.
The manufacturer's detector therefore requires spectral information, supplemented by source information where appropriate.
Vacuum speed is a shared property rather than a spectrum-identification measurement.
218. An experimental team compares the generation and detection of three electromagnetic disturbances:
Arrangement P contains a capacitor being charged by a time-varying current. A magnetic-field measurement near the capacitor gap detects the contribution associated with changing electric flux. Arrangement Q contains an oscillating transmitting aerial whose emitted radio-frequency electromagnetic signal is received across vacuum. Arrangement R contains an X-ray tube where energetic electrons strike a metal target and produce radiation detected beyond the target.
Which statement correctly integrates the physical mechanisms in these arrangements?
ⓐ. P proves that electrons must cross every capacitor gap, while Q and R show that electrons travel from each source to its detector.
ⓑ. P and Q can involve changing electric and magnetic fields, but R cannot produce electromagnetic radiation because its source contains material electrons.
ⓒ. P demonstrates magnetic circulation from changing electric flux; Q demonstrates wireless electromagnetic propagation; R distinguishes electron acceleration from emitted radiation.
ⓓ. P, Q and R all require the emitted radiation to have the same frequency because they involve electromagnetic fields.
Correct Answer: P demonstrates magnetic circulation from changing electric flux; Q demonstrates wireless electromagnetic propagation; R distinguishes electron acceleration from emitted radiation.
Explanation: Arrangement P concerns Maxwell's displacement-current correction to Ampere's circuital law.
A changing electric flux contributes to the circulation of the magnetic field.
This contribution is described by the displacement-current term
\[
I_d=\varepsilon_0\frac{d\Phi_E}{dt}.
\]
The presence of displacement current does not require electrons to cross the insulating capacitor gap.
It accounts for a changing-field contribution that is essential for a consistent electromagnetic description of the charging arrangement.
Arrangement Q involves an oscillating aerial containing accelerating charges.
These charges can generate electromagnetic radiation.
The emitted radio-frequency disturbance contains linked time-varying electric and magnetic fields.
It can propagate through vacuum without requiring a conducting wire between transmitter and receiver.
The receiver responds to the arriving electromagnetic fields rather than collecting the source's original electrons.
Arrangement R uses energetic electrons striking a metal target.
Their interaction with the target can produce X-rays.
The emitted X-rays are electromagnetic radiation and are distinct from the original charged electron beam.
Thus, source charges and emitted electromagnetic waves must not be confused.
The three arrangements illustrate related electromagnetic principles at different stages: changing fields, radiation propagation and radiation production.
They do not require identical radiation frequencies.
Nor do they require free charge carriers to travel continuously through every region occupied by the electromagnetic disturbance.
The correct interpretation connects Maxwell's changing-field contribution with electromagnetic-wave generation and propagation while preserving the distinction between source particles and emitted radiation.
219. A gamma-ray source is placed behind a shutter that can be closed rapidly. At \(t=0\), the shutter closes immediately after the final emitted gamma-ray pulse has passed through it. Two detectors, P and Q, are placed along the same straight vacuum path at distances of \(150\,\text{m}\) and \(450\,\text{m}\), respectively, from the shutter. Assume both detectors respond without instrumental delay and the pulse propagates at \(3.0\times10^8\,\text{m s}^{-1}\). Which statement correctly describes the detection of this final pulse?
ⓐ. Neither detector can record the pulse after \(t=0\), because closing the shutter immediately stops electromagnetic radiation everywhere.
ⓑ. P records the pulse at \(1.5\,\mu\text{s}\), while Q records it at \(0.5\,\mu\text{s}\), because gamma rays travel towards their source.
ⓒ. Both detectors record the pulse simultaneously because gamma radiation has the same vacuum speed as visible light.
ⓓ. P records the pulse at \(0.5\,\mu\text{s}\), while Q records it at \(1.5\,\mu\text{s}\), even though the shutter has already closed.
Correct Answer: P records the pulse at \(0.5\,\mu\text{s}\), while Q records it at \(1.5\,\mu\text{s}\), even though the shutter has already closed.
Explanation: \(\textbf{Interpret the shutter condition:}\)
The final gamma-ray pulse has already passed through the shutter when it closes.
Therefore, the shutter prevents subsequent radiation from entering the detector path but does not recall radiation already propagating through vacuum.
The emitted gamma radiation is electromagnetic radiation.
Its speed in vacuum is
\[
c=3.0\times10^8\,\text{m s}^{-1}
\]
\(\textbf{Determine arrival at Detector P:}\)
The first detector is located at
\[
d_P=150\,\text{m}
\]
The propagation time is
\[
t_P=\frac{d_P}{c}
\]
Substituting,
\[
t_P=\frac{150}{3.0\times10^8}
\]
\[
t_P=5.0\times10^{-7}\,\text{s}
\]
\[
t_P=0.5\,\mu\text{s}
\]
\(\textbf{Determine arrival at Detector Q:}\)
The second detector is located at
\[
d_Q=450\,\text{m}
\]
Therefore,
\[
t_Q=\frac{d_Q}{c}
\]
\[
t_Q=\frac{450}{3.0\times10^8}
\]
\[
t_Q=1.5\times10^{-6}\,\text{s}
\]
\[
t_Q=1.5\,\mu\text{s}
\]
\(\textbf{Compare detection times:}\)
The difference between the arrival times is
\[
\Delta t=t_Q-t_P
\]
\[
\Delta t=1.5-0.5
\]
\[
\Delta t=1.0\,\mu\text{s}
\]
\(\textbf{Physical interpretation:}\)
The final pulse travels from the shutter towards P and then Q.
Closing the shutter changes what radiation can enter the path subsequently; it does not alter the motion of the already emitted pulse.
Both detectors can therefore respond after the shutter has closed.
The calculation does not require gamma radiation to travel faster than other electromagnetic waves.
Instead, the different arrival times result entirely from the different propagation distances.
This distinguishes interruption of radiation production or transmission from propagation of radiation already released into space.
\(\textbf{Final answer:}\)
\[
\boxed{(t_P,t_Q,\Delta t)=(0.5\,\mu\text{s},1.5\,\mu\text{s},1.0\,\mu\text{s})}
\]
220. A simplified teaching model compares four ways of directing controlled gamma radiation towards a specified target region. The model records the fraction of incident radiation energy delivered to the target and the fraction delivered to a separate surrounding region. Assume all entries have been measured under comparable conditions.
| Arrangement | Energy delivered to target | Energy delivered to surrounding region |
|---|
| P | \(85\%\) | \(40\%\) |
| Q | \(75\%\) | \(18\%\) |
| R | \(70\%\) | \(12\%\) |
| S | \(90\%\) | \(30\%\) |
For this model only, an acceptable arrangement must deliver at least \(70\%\) to the target and no more than \(20\%\) to the surrounding region. Among acceptable arrangements, the preferred one maximises the ratio of target delivery to surrounding-region delivery. Which arrangement should be selected?
ⓐ. P, because it delivers more energy to the target than either Q or R.
ⓑ. R, because it satisfies both limits and gives the greatest target-to-surrounding delivery ratio.
ⓒ. Q, because its target delivery is greater than R's, so its ratio must also be greater.
ⓓ. S, because it delivers the greatest fraction of energy to the target.
Correct Answer: R, because it satisfies both limits and gives the greatest target-to-surrounding delivery ratio.
Explanation: \(\textbf{Apply the two acceptance conditions:}\)
The model requires a target-delivery percentage of at least
\[
70\%
\]
It also requires a surrounding-region percentage of no more than
\[
20\%
\]
Both conditions must hold simultaneously.
Arrangement P delivers \(85\%\) to the target, but its surrounding-region value of \(40\%\) exceeds the allowed limit.
Therefore, P is rejected.
Arrangement S similarly delivers a high target fraction but has \(30\%\) surrounding-region delivery.
Therefore, S is also rejected.
Arrangements Q and R satisfy both stated limits.
\(\textbf{Compare the two acceptable arrangements:}\)
The requested comparison is the ratio
\[
K=\frac{\text{target-delivery percentage}}{\text{surrounding-delivery percentage}}
\]
For Q,
\[
K_Q=\frac{75}{18}
\]
\[
K_Q\approx4.17
\]
For R,
\[
K_R=\frac{70}{12}
\]
\[
K_R\approx5.83
\]
Therefore,
\[
K_R\gt K_Q
\]
\(\textbf{Interpret the result:}\)
Q delivers a slightly greater target fraction, but R delivers substantially less energy to the surrounding region.
The smaller surrounding-region value gives R the larger ratio.
It would be incorrect to select an arrangement from target delivery alone without applying the stated exposure constraint.
The reasoning illustrates why controlled gamma-radiation applications must account for interactions with surrounding healthy material as well as the intended target.
Gamma radiation does not distinguish malignant and healthy tissue automatically.
The question is a simplified comparison of supplied measurements, not a real clinical treatment prescription.
The stipulated constraints and preference uniquely select R.
