301. In a crystal diffraction experiment, the first-order maximum for electrons occurs when \(2d\sin\theta=\lambda\). If the accelerating voltage is increased, the value of \(\lambda\) decreases, so the new Bragg angle generally
ⓐ. decreases for same planes and order
ⓑ. increases to \(90^\circ\) always
ⓒ. remains unchanged for all voltages
ⓓ. becomes unrelated to crystal spacing
Correct Answer: decreases for same planes and order
Explanation: For the same crystal planes, \(d\) is fixed. For the same order \(n=1\), Bragg’s relation is \(2d\sin\theta=\lambda\). Increasing accelerating voltage increases electron momentum and decreases the de Broglie wavelength. A smaller \(\lambda\) requires a smaller value of \(\sin\theta\). Therefore, the Bragg angle decreases for the same plane spacing and order.
302. A crystal has plane spacing \(d=2.0\,\text{\AA}\). An electron beam of wavelength \(1.0\,\text{\AA}\) gives first-order constructive scattering. The Bragg angle is
ⓐ. \(30^\circ\)
ⓑ. \(45^\circ\)
ⓒ. \(15^\circ\)
ⓓ. \(60^\circ\)
Correct Answer: \(15^\circ\)
Explanation: \( \textbf{Plane spacing:} \) \(d=2.0\,\text{\AA}\).
\( \textbf{Electron wavelength:} \) \(\lambda=1.0\,\text{\AA}\).
\( \textbf{Order:} \) \(n=1\).
\( \textbf{Bragg relation:} \)
\[
2d\sin\theta=n\lambda
\]
\( \textbf{Substitute known values:} \)
\[
2(2.0)\sin\theta=1.0
\]
\( \textbf{Solve for sine:} \)
\[
\sin\theta=\frac{1.0}{4.0}=0.25
\]
\( \textbf{Angle estimate:} \)
\[
\theta\approx15^\circ
\]
\( \textbf{Final answer:} \) The Bragg angle is about \(15^\circ\), using the angle with the crystal planes in Bragg’s relation.
303. A summary of wave-particle evidence is shown below.
| Observation | Supported idea |
|---|
| P. Photoelectric emission with threshold frequency | Particle nature of radiation |
| Q. Electron diffraction by a crystal | Wave nature of matter |
| R. Interference of light | Wave nature of radiation |
| S. Electron current in a wire | Photon momentum directly |
The mismatched row is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Row P is valid because the threshold-frequency behaviour of the photoelectric effect supports photons and particle-like energy transfer. Row Q is valid because diffraction of electrons supports matter waves. Row R is valid because interference is a wave phenomenon of light. Row S is mismatched because an electron current in a wire does not directly demonstrate photon momentum. Dual-nature evidence must be tied to the specific observed phenomenon.
304. A final comparison says, “Photoelectric effect and electron diffraction prove the same side of dual nature.” The better comparison is that
ⓐ. both prove only the wave nature of radiation
ⓑ. photoelectric effect supports particle nature of radiation, while electron diffraction supports wave nature of matter
ⓒ. both prove that electrons have no particle behaviour
ⓓ. photoelectric effect supports wave nature of matter, while electron diffraction supports particle nature of radiation
Correct Answer: photoelectric effect supports particle nature of radiation, while electron diffraction supports wave nature of matter
Explanation: The photoelectric effect is explained by photons transferring energy \(h\nu\) to electrons. This supports the particle nature of radiation. Electron diffraction shows that moving electrons can produce a wave-like diffraction pattern. This supports the wave nature of matter. Together, the two results show the central idea that radiation and matter both require wave and particle descriptions in suitable phenomena.
305. A moving electron and a photon are both assigned wavelength \( \lambda \). Their common relation is that the magnitude of momentum is
ⓐ. \(p=h\lambda\)
ⓑ. \(p=\lambda h^2\)
ⓒ. \(p=\frac{\lambda}{h}\)
ⓓ. \(p=\frac{h}{\lambda}\)
Correct Answer: \(p=\frac{h}{\lambda}\)
Explanation: For a photon, the momentum relation is \(p=\frac{h}{\lambda}\). For a moving material particle, de Broglie’s relation \( \lambda=\frac{h}{p} \) gives the same momentum form \(p=\frac{h}{\lambda}\). The physical meanings differ because one case involves radiation and the other involves matter. Still, the wavelength-momentum connection has the same inverse form. This shared relation is one reason wave-particle duality appears as a unified idea.
306. A claim says, “Since electrons show diffraction, they should always be treated only as waves.” The most accurate response is that
ⓐ. electrons stop having charge in all wave experiments
ⓑ. electrons become photons whenever their wavelength is measured
ⓒ. diffraction proves that momentum is zero
ⓓ. diffraction shows waves; detection shows particles
Correct Answer: diffraction shows waves; detection shows particles
Explanation: Electron diffraction requires a wave description because it involves interference and maxima at certain directions. However, electrons are still counted and detected as individual charged particles. The wave description and particle description are complementary ways of describing different observations. A diffraction pattern builds up from many localized detections. Dual nature does not replace one description permanently with the other; it selects the useful description according to the experiment.
307. The condition \(n\lambda=2d\sin\theta\) in electron diffraction predicts strong peaks only at certain angles because
ⓐ. electrons lose their charge at every other angle
ⓑ. the crystal spacing becomes zero except at one detector position
ⓒ. path difference is an integral number of wavelengths
ⓓ. photon energy is converted directly into electron rest mass
Correct Answer: path difference is an integral number of wavelengths
Explanation: In crystal diffraction, waves scattered from different planes can interfere with one another. A strong intensity maximum occurs when the path difference equals \(n\lambda\), where \(n\) is an integer. For crystal planes separated by \(d\) and glancing angle \(\theta\), this condition becomes \(n\lambda=2d\sin\theta\). Electrons can satisfy this condition because a moving electron has a de Broglie wavelength. The peak positions therefore support wave-like behaviour of the electron beam rather than a purely classical particle path.
308. In dual-nature physics, the most complete statement among the following is
ⓐ. radiation has only particle nature and matter has only wave nature
ⓑ. all macroscopic objects show easily visible diffraction in daily life
ⓒ. description depends on the observed phenomenon
ⓓ. photoelectric current is unrelated to photon energy or intensity
Correct Answer: description depends on the observed phenomenon
Explanation: Dual nature does not mean that one description is always enough. Light shows wave behaviour in interference and diffraction, but photon behaviour in the photoelectric effect. Matter shows particle behaviour in ordinary detection, but microscopic particles such as electrons can show diffraction. The relevant description depends on the experiment and scale. This is why photoelectric effect, photon relations, de Broglie wavelength, and electron diffraction are connected as parts of one framework.
309. Heisenberg’s uncertainty idea is connected with wave-particle duality because a microscopic particle described by a wave packet cannot generally have
ⓐ. both charge and mass measured in any experiment
ⓑ. both wavelength and frequency related to each other
ⓒ. both kinetic energy and potential difference connected by a formula
ⓓ. unlimited simultaneous precision in position and momentum
Correct Answer: unlimited simultaneous precision in position and momentum
Explanation: A matter wave associated with a moving particle is not like a sharp classical point moving along an exactly known path. To localize a particle more sharply, a wave packet must involve a spread of wavelengths and momenta. This creates a limit on the simultaneous precision of position and momentum. The idea is qualitative here and follows naturally from the wave description of matter. It does not mean charge or mass cannot be measured; it means classical exactness of position and momentum together is not valid at microscopic scales.
310. A claim says, “The uncertainty principle is important only because instruments are poor.” The better statement is that the uncertainty is
ⓐ. removable by using brighter light of any frequency
ⓑ. caused only by air resistance on electrons
ⓒ. quantum limit from wave-particle duality
ⓓ. absent for all microscopic particles
Correct Answer: quantum limit from wave-particle duality
Explanation: The uncertainty principle is not simply a statement about faulty apparatus. It is connected with the wave nature of matter and the way microscopic particles are described. A sharply localized wave packet requires a spread of wavelengths and therefore a spread of momenta. Better instruments can reduce ordinary measurement errors, but they cannot remove this quantum limitation completely. The principle becomes important for microscopic particles, while everyday bodies do not show noticeable effects because their de Broglie wavelengths are extremely small.
311. Consider the following statements about matter waves and uncertainty.
Statement I: Wave-particle duality suggests that microscopic particles cannot always be described by exact classical paths.
Statement II: Matter waves require a material medium like sound waves.
Statement III: A highly localized matter wave is associated with a spread in momentum values.
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I, II and III
ⓓ. I and III only
Correct Answer: I and III only
Explanation: Statement I is true because matter-wave behaviour changes the classical picture of a particle following a perfectly definite path. Statement III is also true at a qualitative level because a narrow wave packet involves a range of wavelengths and momenta. Statement II is false because matter waves are not ordinary mechanical waves travelling through a material medium. They are quantum waves associated with particles. This distinction prevents confusing electron matter waves with sound or water waves.
312. An electron microscope can have better resolving power than an ordinary optical microscope mainly because accelerated electrons can have
ⓐ. wavelengths smaller than visible light
ⓑ. zero momentum at high accelerating voltage
ⓒ. wavelengths larger than radio waves
ⓓ. no wave nature after acceleration
Correct Answer: wavelengths smaller than visible light
Explanation: Resolving fine details requires a probing wavelength small enough to interact with small structures. Visible light has wavelengths of hundreds of \(nm\). Accelerated electrons can have de Broglie wavelengths of the order of angstroms or even smaller under suitable conditions. A smaller wavelength can reveal smaller details. The electron microscope uses the wave nature of electrons; it does not work by making electron momentum zero.
313. An electron microscope uses electrons accelerated through a high potential difference. If the accelerating voltage is increased, the resolving ability can improve because
ⓐ. the electron de Broglie wavelength decreases
ⓑ. the electron charge becomes larger
ⓒ. the electron wavelength becomes equal to visible light wavelength
ⓓ. the electron mass becomes zero
Correct Answer: the electron de Broglie wavelength decreases
Explanation: For non-relativistic accelerated electrons, the de Broglie wavelength follows \(\lambda\propto\frac{1}{\sqrt{V}}\). Increasing the accelerating voltage increases the electron kinetic energy and momentum. Since \(\lambda=\frac{h}{p}\), larger momentum gives a smaller wavelength. Smaller wavelength helps in resolving finer structures. The improvement comes from the wave nature and momentum of electrons, not from changing their charge or rest mass.
314. A comparison of probes used to study small structures is shown below.
| Probe | Relevant wavelength idea | Possible use |
|---|
| P. Visible light | Wavelength of hundreds of \(nm\) | Ordinary optical microscopy |
| Q. Accelerated electrons | de Broglie wavelength can be much smaller | Electron microscopy |
| R. Neutrons | de Broglie wavelength can be comparable with crystal spacings | Crystal studies |
| S. Macroscopic ball | Large easily visible de Broglie wavelength | Routine diffraction grating probe |
The row that should be rejected is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Visible light is suitable for ordinary optical microscopy, but its wavelength limits the smallest details it can resolve. Accelerated electrons can have much smaller de Broglie wavelengths, which is why electron microscopy is possible. Neutrons can also show matter-wave diffraction and are useful in crystal studies when their wavelengths are suitable. A macroscopic ball has an extremely tiny de Broglie wavelength and is not used as a routine wave probe in this sense. Row S reverses the macroscopic-scale limitation of matter waves.
315. A neutron beam is used to study a crystal. The use of neutrons in diffraction studies is possible because neutrons
ⓐ. must become photons before entering the crystal
ⓑ. moving neutrons have de Broglie wavelength
ⓒ. have no momentum because they are uncharged
ⓓ. produce photoelectric current directly in every metal
Correct Answer: moving neutrons have de Broglie wavelength
Explanation: de Broglie wavelength is associated with momentum, not with electric charge alone. A neutron is electrically neutral, but a moving neutron still has momentum. Therefore, it can have a matter wavelength \( \lambda=\frac{h}{p} \). If that wavelength is comparable with crystal spacing, diffraction can occur. Electric neutrality changes how the neutron interacts with electric fields, but it does not remove its wave nature.
316. A laboratory uses electrons accelerated through \(10^4\,V\). Using \(\lambda=\frac{12.27}{\sqrt{V}}\,\text{\AA}\), the electron wavelength is closest to
ⓐ. \(0.123\,\text{\AA}\)
ⓑ. \(1.23\,\text{\AA}\)
ⓒ. \(12.27\,\text{\AA}\)
ⓓ. \(122.7\,\text{\AA}\)
Correct Answer: \(0.123\,\text{\AA}\)
Explanation: \( \textbf{Accelerating potential:} \) \(V=10^4\,V\).
\( \textbf{Electron wavelength relation:} \)
\[
\lambda=\frac{12.27}{\sqrt{V}}\,\text{\AA}
\]
\( \textbf{Square root of voltage:} \)
\[
\sqrt{10^4}=100
\]
\( \textbf{Substitution:} \)
\[
\lambda=\frac{12.27}{100}\,\text{\AA}
\]
\( \textbf{Calculation:} \)
\[
\lambda=0.1227\,\text{\AA}
\]
\( \textbf{Rounded value:} \)
\[
\lambda\approx0.123\,\text{\AA}
\]
\( \textbf{Scale interpretation:} \) This wavelength is far smaller than visible-light wavelengths.
\( \textbf{Final answer:} \) The accelerated electron wavelength is about \(0.123\,\text{\AA}\), supporting its use as a short-wavelength probe.
317. A photon and an electron are assigned the same wavelength \( \lambda \). Which comparison is correct?
ⓐ. They have the same rest mass because their wavelengths are the same
ⓑ. They must travel with the same speed \(c\) in vacuum
ⓒ. same momentum magnitude, different energy relations
ⓓ. They must have the same kinetic energy \(\frac{hc}{\lambda}\)
Correct Answer: same momentum magnitude, different energy relations
Explanation: For a photon, \(p=\frac{h}{\lambda}\). For an electron matter wave, \(\lambda=\frac{h}{p}\), so \(p=\frac{h}{\lambda}\) as well. Same wavelength therefore gives the same momentum magnitude. However, a photon has energy \(E=\frac{hc}{\lambda}\), while a non-relativistic electron with the same momentum has kinetic energy \(K=\frac{p^2}{2m_e}\). Equal wavelength does not make a photon and an electron identical physical objects.
318. A photon has wavelength \(1.0\,\text{\AA}\). An electron also has de Broglie wavelength \(1.0\,\text{\AA}\). The photon energy is \(12.4\,keV\), while the electron kinetic energy is much smaller. The main reason is that
ⓐ. photon and electron energy formulas differ
ⓑ. the electron has no momentum at wavelength \(1.0\,\text{\AA}\)
ⓒ. the photon has electron rest mass
ⓓ. equal wavelength always means equal energy for all objects
Correct Answer: photon and electron energy formulas differ
Explanation: A photon of wavelength \( \lambda \) has energy \(E=\frac{hc}{\lambda}\). An electron with de Broglie wavelength \( \lambda \) has momentum \(p=\frac{h}{\lambda}\). For a non-relativistic electron, kinetic energy is \(K=\frac{p^2}{2m_e}\). Thus the same wavelength gives the same momentum magnitude relation, but not the same energy formula. The formula used must match whether the object is a photon or a massive particle.
319. For an electron of de Broglie wavelength \(1.0\,\text{\AA}\), using \(\lambda=\frac{12.27}{\sqrt{V}}\,\text{\AA}\), the accelerating voltage is about \(151\,V\). What is the kinetic energy of the electron in \(eV\)?
ⓐ. \(1.0\,eV\)
ⓑ. \(151\,eV\)
ⓒ. \(12.27\,eV\)
ⓓ. \(1240\,eV\)
Correct Answer: \(151\,eV\)
Explanation: \( \textbf{Given wavelength:} \) \(\lambda=1.0\,\text{\AA}\).
\( \textbf{Electron voltage relation:} \)
\[
\lambda=\frac{12.27}{\sqrt{V}}\,\text{\AA}
\]
\( \textbf{From the relation:} \)
\[
V\approx151\,V
\]
\( \textbf{Energy gained by an electron:} \)
\[
K=eV
\]
\( \textbf{Electron-volt meaning:} \) An electron accelerated through \(1\,V\) gains \(1\,eV\).
\( \textbf{Therefore:} \)
\[
K=151\,eV
\]
\( \textbf{Comparison note:} \) This is not the same as a \(1.0\,\text{\AA}\) photon energy, which would be \(12.4\,keV\).
\( \textbf{Final answer:} \) The electron kinetic energy is about \(151\,eV\).
320. A student uses \(E=\frac{hc}{\lambda}\) to find the kinetic energy of an electron from its de Broglie wavelength. The error is that this formula gives
ⓐ. electron momentum directly
ⓑ. work function of every metal
ⓒ. stopping potential for every surface
ⓓ. photon energy, not electron \(K\)
Correct Answer: photon energy, not electron \(K\)
Explanation: The formula \(E=\frac{hc}{\lambda}\) is the energy of a photon. An electron with de Broglie wavelength \( \lambda \) is a massive particle, so its wavelength first gives momentum through \(p=\frac{h}{\lambda}\). Its non-relativistic kinetic energy is then found from \(K=\frac{p^2}{2m_e}\), or from \(K=eV\) if it was accelerated through a known potential. Using the photon formula for an electron overestimates the energy. The same symbol \( \lambda \) appears in both formulas, but the physical object decides which relation applies.
