401. Electron capture in a nucleus can be represented as:
ⓐ. \(p+e^-\rightarrow n+\nu\)
ⓑ. \(n\rightarrow p+e^-+\bar{\nu}\)
ⓒ. \(p\rightarrow n+e^++\nu\)
ⓓ. \({}^{4}_{2}He\rightarrow2p+2n\)
Correct Answer: \(p+e^-\rightarrow n+\nu\)
Explanation: In electron capture, an inner atomic electron is captured by the nucleus. A proton combines with this electron and changes into a neutron. A neutrino is emitted in the process. The nuclear result is that \(Z\) decreases by \(1\), while \(A\) remains unchanged. This daughter change is similar to beta plus decay, although the emitted particle pattern is different. The relation \(n\rightarrow p+e^-+\bar{\nu}\) describes beta minus decay instead.
402. The nuclear equation \({}^{40}_{19}K+{}^{0}_{-1}e\rightarrow{}^{40}_{18}Ar+\nu\) represents:
ⓐ. Alpha decay
ⓑ. Beta minus decay
ⓒ. Electron capture
ⓓ. Gamma decay
Correct Answer: Electron capture
Explanation: The left side includes an electron \({}^{0}_{-1}e\) being captured by the nucleus. The mass number remains \(40\), so the total nucleon count is unchanged. The atomic number changes from \(19\) to \(18\), meaning one proton has effectively changed into a neutron. This is the signature of electron capture. Alpha decay would reduce \(A\) by \(4\) and \(Z\) by \(2\). Gamma decay would not change \(A\) or \(Z\).
403. A decay chain changes a parent \({}^{A}_{Z}X\) into a final nucleus with \(A-12\) and \(Z-4\). If only alpha and beta minus decays occur, the number of alpha and beta minus decays is:
ⓐ. \(3\) alpha decays and \(2\) beta minus decays
ⓑ. \(2\) alpha decays and \(3\) beta minus decays
ⓒ. \(3\) alpha decays and \(0\) beta minus decays
ⓓ. \(1\) alpha decay and \(2\) beta minus decays
Correct Answer: \(3\) alpha decays and \(2\) beta minus decays
Explanation: \( \textbf{Mass-number change:} \)
\[
\Delta A=-12
\]
\( \textbf{Alpha decay effect:} \) Each alpha decay gives \(\Delta A=-4\).
\( \textbf{Number of alpha decays:} \)
\[
\frac{12}{4}=3
\]
\( \textbf{Atomic-number change from \(3\) alpha decays:} \)
\[
\Delta Z=3(-2)=-6
\]
\( \textbf{Required final atomic-number change:} \)
\[
\Delta Z=-4
\]
\( \textbf{Needed increase after alpha decays:} \)
\[
-4-(-6)=+2
\]
\( \textbf{Beta minus effect:} \) Each beta minus decay gives \(\Delta Z=+1\) and \(\Delta A=0\).
\( \textbf{Final answer:} \) The chain requires \(3\) alpha decays and \(2\) beta minus decays. The alpha count is fixed first by the mass-number change.
404. A nuclide emits two beta minus particles and one alpha particle in any order. The net change in \(A\) and \(Z\) is:
ⓐ. \(\Delta A=-4,\ \Delta Z=-4\)
ⓑ. \(\Delta A=-4,\ \Delta Z=0\)
ⓒ. \(\Delta A=0,\ \Delta Z=+2\)
ⓓ. \(\Delta A=-8,\ \Delta Z=0\)
Correct Answer: \(\Delta A=-4,\ \Delta Z=0\)
Explanation: One alpha decay changes \(A\) by \(-4\) and \(Z\) by \(-2\). Each beta minus decay changes \(A\) by \(0\) and \(Z\) by \(+1\). Two beta minus decays therefore give a total \(\Delta Z=+2\). Combining the changes gives \(\Delta A=-4\) and \(\Delta Z=-2+2=0\). The order of emissions does not change the net result. The final nuclide has the same atomic number as the initial nuclide but a mass number smaller by \(4\).
405. A reaction is written in the form \(a+X\rightarrow Y+b\). In this notation:
ⓐ. \(a\) is always a gamma ray, and \(X\) is always an electron
ⓑ. \(Y\) is the target, and \(b\) is the incident projectile
ⓒ. Projectile, target, product, emitted particle
ⓓ. \(X\) is the energy released, and \(a\) is the half-life
Correct Answer: Projectile, target, product, emitted particle
Explanation: Nuclear reaction notation often writes an incoming projectile \(a\) striking a target nucleus \(X\). The reaction produces a product nucleus \(Y\) and an emitted particle \(b\). This form helps track which particle enters and which particle leaves. Conservation of mass number and atomic number must still be applied to the full equation. The symbols are general and are not restricted to gamma rays or electrons. The notation describes a nuclear transformation, not a decay constant or half-life.
406. A reaction \(a+{}^{27}_{13}Al\rightarrow{}^{30}_{15}P+{}^{1}_{0}n\) is balanced. The projectile \(a\) is:
ⓐ. \({}^{2}_{1}H\)
ⓑ. \({}^{1}_{1}H\)
ⓒ. \({}^{0}_{-1}e\)
ⓓ. \({}^{4}_{2}He\)
Correct Answer: \({}^{4}_{2}He\)
Explanation: \( \textbf{Balance mass number:} \)
\[
A_a+27=30+1
\]
\[
A_a+27=31
\]
\[
A_a=4
\]
\( \textbf{Balance atomic number:} \)
\[
Z_a+13=15+0
\]
\[
Z_a=2
\]
\( \textbf{Projectile identity:} \)
\[
a={}^{4}_{2}He
\]
\( \textbf{Final answer:} \) The projectile is \({}^{4}_{2}He\), an alpha particle. The emitted neutron contributes mass number \(1\) but no charge number.
407. A gamma ray and an X-ray may both be electromagnetic radiation. For nuclear processes, the main distinction is that:
ⓐ. Gamma: nuclear; X-rays: atomic electronic
ⓑ. Gamma rays have positive charge, while X-rays have negative charge
ⓒ. Gamma rays have mass number \(4\), while X-rays have mass number \(0\)
ⓓ. Gamma rays are always slower than alpha particles
Correct Answer: Gamma: nuclear; X-rays: atomic electronic
Explanation: Both gamma rays and X-rays are electromagnetic radiation. The distinction is mainly their origin. Gamma rays in radioactivity come from changes in nuclear energy states. X-rays are commonly produced by electronic transitions or deceleration of fast electrons outside the nucleus. Both photons have no charge and no mass number. The nuclear origin of gamma rays is why gamma emission can follow alpha or beta decay when the daughter nucleus is left excited.
408. A source emits radiation that causes strong ionisation in a very short path. A second source emits radiation that is detected after passing through thick shielding more easily. The best comparison is:
ⓐ. The first is likely gamma radiation, while the second is likely alpha radiation
ⓑ. The first is likely alpha radiation, while the second is likely gamma radiation
ⓒ. Both must be beta minus radiation
ⓓ. Both must be ordinary visible light
Correct Answer: The first is likely alpha radiation, while the second is likely gamma radiation
Explanation: Alpha particles have high ionising power because they are massive and carry charge \(+2e\). They lose energy quickly and therefore travel only a short distance in matter. Gamma rays are neutral high-energy photons and have much greater penetrating power. They can pass through shielding more effectively than alpha particles. The two descriptions compare ionisation and penetration, which are different radiation properties. Strong ionisation over short range points to alpha radiation, while high penetration points to gamma radiation.
409. A fission fragment pair is not unique for \({}^{235}_{92}U\) fission. This means:
ⓐ. Conservation of mass number is not required in fission
ⓑ. Different fragment pairs can satisfy conservation
ⓒ. Every fission must produce exactly the same daughter nuclei
ⓓ. Fission products are always identical to the parent nucleus
Correct Answer: Different fragment pairs can satisfy conservation
Explanation: Fission of a heavy nucleus can produce different pairs of medium-mass fragments. It often also emits neutrons and gamma radiation. Even though the fragment pair can vary, each possible fission event must conserve total mass number and charge number. The energy released can also vary somewhat depending on the product masses and excitation states. Therefore, fission is not represented by only one possible fragment pair in all events. The variety of products does not remove the need for nuclear bookkeeping.
410. A fission equation is proposed:
\[
{}^{235}_{92}U+{}^{1}_{0}n\rightarrow{}^{140}_{54}Xe+{}^{94}_{38}Sr+x{}^{1}_{0}n
\]
The value of \(x\) required by mass-number conservation is:
ⓐ. \(3\)
ⓑ. \(4\)
ⓒ. \(1\)
ⓓ. \(2\)
Correct Answer: \(2\)
Explanation: \( \textbf{Left-side mass number:} \)
\[
235+1=236
\]
\( \textbf{Right-side mass number without emitted neutrons:} \)
\[
140+94=234
\]
\( \textbf{Mass number carried by emitted neutrons:} \)
\[
x(1)=x
\]
\( \textbf{Mass balance:} \)
\[
236=234+x
\]
\( \textbf{Solve:} \)
\[
x=2
\]
\( \textbf{Charge check:} \)
\[
92=54+38
\]
\( \textbf{Final answer:} \) The value of \(x\) is \(2\). The emitted neutrons are needed to balance the total nucleon count.
411. A deuterium-tritium fusion reaction releases \(17.6\,\text{MeV}\) per event. Using \(1\,\text{MeV}=1.6\times10^{-13}\,\text{J}\), the energy released in \(1.0\times10^{20}\) such reactions is closest to:
ⓐ. \(2.82\times10^{-12}\,\text{J}\)
ⓑ. \(1.10\times10^{21}\,\text{J}\)
ⓒ. \(1.76\times10^{21}\,\text{J}\)
ⓓ. \(2.82\times10^8\,\text{J}\)
Correct Answer: \(2.82\times10^8\,\text{J}\)
Explanation: \( \textbf{Energy per reaction:} \)
\[
E_f=17.6\,\text{MeV}
\]
\( \textbf{Convert to joules:} \)
\[
E_f=17.6(1.6\times10^{-13})\,\text{J}
\]
\[
E_f=28.16\times10^{-13}\,\text{J}=2.816\times10^{-12}\,\text{J}
\]
\( \textbf{Number of reactions:} \)
\[
N=1.0\times10^{20}
\]
\( \textbf{Total energy:} \)
\[
E=NE_f
\]
\[
E=(1.0\times10^{20})(2.816\times10^{-12})\,\text{J}
\]
\( \textbf{Calculation:} \)
\[
E=2.816\times10^8\,\text{J}
\]
\( \textbf{Final answer:} \) The released energy is about \(2.82\times10^8\,\text{J}\). The conversion must be applied per reaction before multiplying by the number of reactions.
412. A fusion plasma must be confined for a useful time because:
ⓐ. Confinement keeps hot nuclei colliding
ⓑ. Fusion requires nuclei to be cooled to ordinary room temperature
ⓒ. Confinement changes protons into electrons before fusion
ⓓ. Fusion energy comes from chemical burning of the container
Correct Answer: Confinement keeps hot nuclei colliding
Explanation: Fusion requires light positive nuclei to come very close despite Coulomb repulsion. Very high temperature gives nuclei high kinetic energy, but the hot plasma also tends to expand and lose energy. Confinement keeps the nuclei together long enough for enough close collisions to occur. In stars, gravity provides this confinement on a huge scale. In laboratory fusion, confinement is a major practical difficulty. The need for confinement comes from maintaining fusion conditions, not from chemical burning or converting protons into electrons.
413. A point-like gamma source is observed from farther away while shielding and source strength remain unchanged. Increasing distance reduces exposure mainly because:
ⓐ. The gamma photons lose all charge
ⓑ. Larger-area spreading
ⓒ. The half-life becomes shorter
ⓓ. The nuclei stop decaying
Correct Answer: Larger-area spreading
Explanation: Radiation emitted from a small source spreads out as it travels. At a larger distance, the same emitted radiation is distributed over a larger area. Therefore, the intensity reaching a small detector or body region is reduced. This is why increasing distance is a basic radiation-safety method along with reducing time and using shielding. The source nuclei continue to decay according to their half-life. The effect of distance is geometrical spreading, not a change in nuclear decay constant.
414. A source has activity \(A_c=5.0\times10^6\,\text{Bq}\). The average number of decays in \(2.0\,\text{min}\) is:
ⓐ. \(6.0\times10^8\)
ⓑ. \(1.0\times10^7\)
ⓒ. \(2.5\times10^6\)
ⓓ. \(1.5\times10^5\)
Correct Answer: \(6.0\times10^8\)
Explanation: \( \textbf{Activity:} \)
\[
A_c=5.0\times10^6\,\text{Bq}=5.0\times10^6\,\text{s}^{-1}
\]
\( \textbf{Time conversion:} \)
\[
2.0\,\text{min}=120\,\text{s}
\]
\( \textbf{Number of decays relation:} \)
\[
\text{decays}=A_c t
\]
\( \textbf{Substitution:} \)
\[
\text{decays}=(5.0\times10^6)(120)
\]
\( \textbf{Calculation:} \)
\[
\text{decays}=6.0\times10^8
\]
\( \textbf{Final answer:} \) The average number of decays is \(6.0\times10^8\). This uses activity as decays per second over a short interval where activity can be treated as nearly constant.
415. A nuclear reaction conserves \(A\) and \(Z\), but its products are left in an excited nuclear state and then emit a gamma ray. The gamma ray mainly carries away:
ⓐ. Four units of mass number
ⓑ. Two units of atomic number
ⓒ. Excitation energy
ⓓ. One orbital electron from the daughter atom
Correct Answer: Excitation energy
Explanation: A nuclear reaction can leave a product nucleus in an excited state. Gamma emission allows that nucleus to move to a lower energy state. The emitted gamma photon carries away energy but no mass number and no charge number. Therefore, it does not change \(A\) or \(Z\). This is different from alpha emission, which carries away nucleons, and beta decay, which changes proton-neutron balance. Gamma emission completes the energy adjustment of the nucleus without changing its identity.
416. A mixed nuclear data record is shown below.
| Initial nucleus | Process | Later measurement |
|---|
| \({}^{A}_{Z}X\) | One alpha decay followed by one gamma emission | Activity halves after time \(T\) |
The daughter after the alpha decay and the meaning of \(T\) are:
ⓐ. \({}^{A}_{Z+1}Y\), and \(T\) is the mean life
ⓑ. \({}^{A-4}_{Z-2}Y\), and \(T\) is half-life
ⓒ. \({}^{A}_{Z}X\), and \(T\) is the nuclear radius
ⓓ. \({}^{A-4}_{Z}Y\), and \(T\) is the binding energy per nucleon
Correct Answer: \({}^{A-4}_{Z-2}Y\), and \(T\) is half-life
Explanation: Alpha decay emits \({}^{4}_{2}He\), so the daughter after alpha emission has mass number \(A-4\) and atomic number \(Z-2\). Gamma emission after that can lower the energy of the daughter nucleus but does not change \(A\) or \(Z\). The later measurement says the activity halves after time \(T\). The time for activity or number of undecayed nuclei to halve is the half-life. Mean life is related but not equal to the repeated halving time. The process combines nuclear equation bookkeeping with radioactive decay terminology.
417. A reaction product has a higher binding energy per nucleon but the reaction has \(Q<0\) according to the complete mass data. The safest conclusion is:
ⓐ. The reaction must release energy because one product has higher \(\frac{B}{A}\)
ⓑ. \(Q<0\) means charge is not conserved
ⓒ. Binding energy per nucleon is always enough without checking all products
ⓓ. The complete mass-energy calculation decides the actual energy balance
Correct Answer: The complete mass-energy calculation decides the actual energy balance
Explanation: Binding energy per nucleon is a useful guide for broad stability trends. However, a reaction energy must be calculated using the complete initial and final mass-energy balance. All products, emitted particles, excitation energies, and kinetic energies must be included where relevant. A single product with high \(\frac{B}{A}\) does not by itself prove the whole reaction releases energy. If the complete mass data give \(Q<0\), the reaction is endoergic. Broad graph reasoning guides expectations, but the \(Q\)-value gives the actual energy sign for the stated reaction.
418. A nucleus near \(A\approx56\) is compared with a very heavy nucleus using the binding-energy curve. The heavy nucleus can release energy by fission more easily than the \(A\approx56\) nucleus because:
ⓐ. The medium nucleus has no protons
ⓑ. The heavy nucleus has zero binding energy
ⓒ. Heavy gains \(B/A\); medium is near the peak
ⓓ. Fission energy does not depend on binding energy at all
Correct Answer: Heavy gains \(B/A\); medium is near the peak
Explanation: The binding energy per nucleon curve has a broad maximum near \(A\approx56\). Nuclei near this region are already strongly bound on average. Very heavy nuclei lie on the slowly falling right side of the curve, so splitting into medium-mass fragments can increase \(\frac{B}{A}\). That increase corresponds to lower final mass-energy and energy release. A nucleus already near the peak has much less opportunity to gain average binding through ordinary fission. The reason is graph-based stability, not absence of protons or zero binding.
419. A reaction releases \(2.4\times10^{12}\,\text{J}\) of energy. The equivalent mass loss is closest to \(\left(c=3.0\times10^8\,\text{m s}^{-1}\right)\):
ⓐ. \(8.0\times10^3\,\text{kg}\)
ⓑ. \(7.2\times10^{20}\,\text{kg}\)
ⓒ. \(2.7\times10^{-5}\,\text{kg}\)
ⓓ. \(2.7\times10^{-8}\,\text{kg}\)
Correct Answer: \(2.7\times10^{-5}\,\text{kg}\)
Explanation: \( \textbf{Energy released:} \)
\[
E=2.4\times10^{12}\,\text{J}
\]
\( \textbf{Mass-energy relation:} \)
\[
E=\Delta mc^2
\]
\( \textbf{Solve for mass loss:} \)
\[
\Delta m=\frac{E}{c^2}
\]
\( \textbf{Square of \(c\):} \)
\[
c^2=(3.0\times10^8)^2=9.0\times10^{16}\,\text{m}^2\text{s}^{-2}
\]
\( \textbf{Substitution:} \)
\[
\Delta m=\frac{2.4\times10^{12}}{9.0\times10^{16}}\,\text{kg}
\]
\( \textbf{Calculation:} \)
\[
\Delta m=0.267\times10^{-4}\,\text{kg}=2.7\times10^{-5}\,\text{kg}
\]
\( \textbf{Final answer:} \) The equivalent mass loss is about \(2.7\times10^{-5}\,\text{kg}\). A small mass decrease can correspond to a large energy release because \(c^2\) is very large.
420. A full nuclei problem asks for the most reliable stability comparison between two nuclides, then asks whether a reaction between them releases energy. The best method is:
ⓐ. Use total binding energy alone for both questions
ⓑ. Use atomic number alone for both questions
ⓒ. Use shielding thickness alone for both questions
ⓓ. Use \(B/A\) and \(Q\) separately
Correct Answer: Use \(B/A\) and \(Q\) separately
Explanation: Binding energy per nucleon \(\frac{B}{A}\) is the usual average measure for comparing nuclear stability across different mass numbers. It avoids the simple size effect that makes large nuclei often have larger total binding energy. A reaction energy question needs the complete mass-energy balance of the stated reactants and products. That balance is expressed by the \(Q\)-value. Total binding energy or atomic number alone cannot answer both questions reliably. Stability comparison and reaction energetics are connected, but they are not identical calculations.
