JEE Physics Dual Nature Of Radiation and Matter MCQs Set 12

Answer: (b) ∞

Explanation:
1. A neutron carries no charge, so an electric potential difference exerts no force on it.
2. The neutron therefore stays at rest: \( p=0 \).
3. \( \lambda=\frac hp\to\infty \).

Teacher's Note:
1. Only charged particles are accelerated by a potential difference.
2. The formulas with V in the options hold for an electron or a charged particle, not a neutron.

Answer: (a) remains constant

Explanation:
1. The magnetic force is always perpendicular to the velocity, so it does no work.
2. The speed, and hence the momentum magnitude, does not change: \( \lambda=\frac hp \) stays constant.

Teacher's Note:
1. Magnetic force changes the direction of motion, not the speed.
2. The de Broglie wavelength depends only on the magnitude of momentum.

Answer: (d) 5

Explanation:
1. Bohr's quantisation condition in wave form: \( 2\pi r_n=n\lambda \).
2. For the fifth orbit n = 5, so five complete wavelengths fit around the orbit.

Teacher's Note:
1. A stable orbit holds a whole number of standing waves.
2. The number of wavelengths equals the principal quantum number n.

Answer: (c) both \( 1.15\times10^{-19}\ \text{N s} \)

Explanation:
1. \( p_\alpha=\frac h\lambda=\frac{6.63\times10^{-34}}{5.76\times10^{-15}}\approx1.15\times10^{-19}\ \text{N s} \).
2. The nucleus was initially at rest, so momentum conservation gives the daughter nucleus an equal and opposite momentum of the same magnitude.

Teacher's Note:
1. Total momentum before decay is zero.
2. The masses are not needed for the momenta, only for the energies.

Answer: (d) 6.2 MeV

Explanation:
1. \( p=\frac h\lambda\approx1.15\times10^{-19}\ \text{kg m/s} \).
2. \( K_\alpha=\frac{p^2}{2m}=\frac{(1.15\times10^{-19})^2}{2\times4.002\times1.66\times10^{-27}}\approx9.97\times10^{-13}\ \text{J} \).
3. Dividing by \( 1.6\times10^{-13} \) J per MeV gives about 6.2 MeV (the recoiling daughter nucleus only gets about 0.11 MeV).

Teacher's Note:
1. Use K = p²/2m with the same momentum for both fragments.
2. Convert amu to kg with 1 amu = 1.66 × 10⁻²⁷ kg.

Answer: (c) \( \sqrt3 \)

Explanation:
1. Kinetic energy in the zero-potential half: 15 eV. Kinetic energy in the 10 eV half: 15 − 10 = 5 eV.
2. \( \lambda\propto\frac1{\sqrt K} \), so \( \frac{\lambda_{10}}{\lambda_0}=\sqrt{\frac{15}{5}}=\sqrt3 \).

Teacher's Note:
1. Find the kinetic energy in each region first.
2. A smaller kinetic energy means a longer wavelength.

Answer: (b) the same

Explanation:
1. The magnetic force \( q\vec v\times\vec B \) is perpendicular to the velocity and does no work.
2. The speed stays \( 10^6 \) m/s, so the momentum magnitude and the wavelength are unchanged.

Teacher's Note:
1. Work done by a magnetic force is zero.
2. Check whether the speed changes, not the direction.

Answer: (a) \( \frac h{\sqrt{fmkT}} \)

Explanation:
1. \( K=\frac{p^2}{2m}=\frac f2kT\Rightarrow p=\sqrt{fmkT} \).
2. \( \lambda=\frac hp=\frac h{\sqrt{fmkT}} \).

Teacher's Note:
1. Write p in terms of the kinetic energy.
2. Then substitute in λ = h/p.

Answer: (b) the kinetic energy of the electrons reaching the collector will increase

Explanation:
1. A downward field exerts an upward force on electrons, towards the collector, so it accelerates them.
2. Saturation current means every emitted electron already reaches the collector, so the current cannot increase.
3. The stopping potential and the threshold wavelength depend only on the light and the emitter material, so they are unchanged.
4. The extra work done by the field raises the kinetic energy with which electrons arrive.

Teacher's Note:
1. Saturation current is limited by the number of emitted electrons, not by the field.
2. Stopping potential and threshold depend on the metal and the light only.

Answer: (d) none of these

Explanation:
1. Blue light has not caused emission, so its frequency is below the threshold frequency of the surface.
2. Red, infrared and yellow light all have lower frequencies than blue light, so none of them can eject electrons.

Teacher's Note:
1. Compare frequencies: red < yellow < blue.
2. Emission needs frequency at least the threshold frequency.

Answer: (c) not come out

Explanation:
1. In photoemission one electron absorbs one photon. The photon energy 1 eV is less than the work function 2 eV.
2. Heat energy is shared among many atoms and cannot add to the photon's energy for the same electron, so the electron does not escape.

Teacher's Note:
1. Photoelectric emission is a one-photon, one-electron process.
2. Intensity only changes the number of photons, not the energy of each.

Answer: (b) are one quarter as numerous

Explanation:
1. Intensity falls as the inverse square of distance, so at 2 m it is one quarter of its earlier value.
2. The frequency of the light is unchanged, so each photoelectron has the same energy and momentum as before.
3. The number of photoelectrons per second is proportional to intensity, so it becomes one quarter.

Teacher's Note:
1. Distance changes intensity, not frequency.
2. Number of electrons is proportional to intensity.

Answer: (a) 0.65 eV

Explanation:
1. Photon energy \( E=13.6\left(\frac1{2^2}-\frac1{4^2}\right)=13.6\times\frac3{16}=2.55\ \text{eV} \).
2. \( K_{max}=h\nu-\phi=2.55-1.9=0.65\ \text{eV} \).

Teacher's Note:
1. First find the energy of the emitted photon.
2. Subtract the work function.

Answer: (d) more than n times the initial kinetic energy

Explanation:
1. \( K=h\nu-\phi \) so \( h\nu=K+\phi \).
2. \( K'=nh\nu-\phi=n(K+\phi)-\phi=nK+(n-1)\phi \).
3. Since \( (n-1)\phi\gt 0 \), K' is more than nK.

Teacher's Note:
1. Write the new energy equation and eliminate hν.
2. The work function is a fixed positive quantity.

Answer: (c) is related to the maximum kinetic energy of the emitted photoelectrons

Explanation:
1. At the stopping potential the work done against the field just stops the fastest electrons: \( eV_0=K_{max} \).
2. Also \( eV_0=h\nu-\phi \), so it depends on the frequency and the metal, but not on intensity.

Teacher's Note:
1. Stopping potential measures the maximum kinetic energy.
2. Intensity affects only the photocurrent.

Answer: (b) the separation between the gold leaves increases

Explanation:
1. Ultraviolet light causes photoelectric emission from the zinc plate; the emitted electrons leave the system.
2. Losing electrons makes the plate and leaves more positively charged, so the leaves repel each other more and diverge further.

Teacher's Note:
1. Photoemission removes negative charge from the plate.
2. More positive charge on the leaves means greater repulsion.

Answer: (b) (i) is true but (ii) is false

Explanation:
1. Photons per second = intensity/(hν). Doubling both leaves the photon rate unchanged, so the saturation current stays about the same: (i) is true.
2. \( K'=2h\nu-\phi\neq2(h\nu-\phi) \); it is more than double, so (ii) is false.

Teacher's Note:
1. The saturation current depends on the number of photons per second.
2. Kmax = hν − φ is not proportional to frequency.

Answer: (a) the photoelectrons are emitted but are reabsorbed by the emitter metal

Explanation:
1. The stopping potential retards the electrons: even the fastest emitted electron is brought to rest just before reaching the collector.
2. Emission continues, but the electrons turn back and return to the emitter, so the net current is zero.

Teacher's Note:
1. Emission depends on the light, not on the applied voltage.
2. The retarding field sends electrons back towards the emitter.

Answer: (a) a and b only

Explanation:
1. Intensity is the energy per unit area per unit time; for light of fixed frequency, more intensity means more photons per second, and so more energy per second: (a) and (b) are true.
2. The energy of each photon (hν) depends only on frequency, and all photons move at speed c, so (c) and (d) are false.

Teacher's Note:
1. Photon energy depends on frequency, not intensity.
2. All photons travel at the speed of light.

Answer: (d) it may come out with kinetic energy less than hν − φ

Explanation:
1. The kinetic energy hν − φ is the maximum value, for electrons at the surface that lose no energy to collisions.
2. An electron may lose some energy on its way out through the metal, so it can emerge with less than hν − φ, or fail to emerge at all.
3. Hence only statement (d) is always true.

Teacher's Note:
1. Kmax is an upper limit for the kinetic energy.
2. Energy losses inside the metal reduce the energy of some electrons.

Answer: (d) \( \text{greater than }\frac83\ \text{V} \)

Explanation:
1. Photon energy at 400 nm: 1240/400 = 3.1 eV, so \( \phi=3.1-2=1.1\ \text{eV} \).
2. At 300 nm: 1240/300 = 4.13 eV, so \( eV_0'=4.13-1.1=3.03\ \text{eV} \).
3. 3.03 V is greater than 8/3 V = 2.67 V.

Teacher's Note:
1. Find the work function from the first case.
2. Stopping potential is not directly proportional to frequency because of the work function.