JEE Physics Dual Nature Of Radiation And Matter MCQs Set 14

Answer: (a) 60

Explanation:
1. Each shell n holds \( 2n^2 \) electrons, i.e. holds that many elements added.
2. Total \( 2(1^2+2^2+3^2+4^2)=2 imes30=60 \).

Teacher's Note:
1. Sum the shell capacities 2, 8, 18, 32.

Answer: (a) \( E/p \)

Explanation:
1. \( \frac Ep=\frac{hf}{h/\lambda}=f\lambda \).
2. Since \( f\lambda=c \) the speed is E/p.

Teacher's Note:
1. For any wave, speed = frequency x wavelength.

Answer: (c) \( n_1/n_2=f_2/f_1 \)

Explanation:
1. Equal total energy: \( n_1hf_1=n_2hf_2 \).
2. Hence \( n_1/n_2=f_2/f_1 \).

Teacher's Note:
1. More photons are needed when each photon carries less energy.

Answer: (d) (i), (iii), (iv)

Explanation:
1. Statements (i), (iii) and (iv) cannot be explained by the wave theory but follow from photons.
2. Statement (ii) is true of charge but it is not evidence for the quantum nature of light.

Teacher's Note:
1. Wave theory predicts no threshold frequency and a time lag.

Answer: (c) more than double

Explanation:
1. Before: \( K=hf-\phi \). After: \( K'=2hf-\phi=2K+\phi \).
2. As \( \phi>0 \), K' is more than 2K.

Teacher's Note:
1. The work function is subtracted only once after doubling the photon energy.

Answer: (b) the stopping potential will be 0.6 V

Explanation:
1. The stopping potential depends on frequency only, so it stays 0.6 V.
2. The saturation current falls with the intensity: \( 18\times\left(\frac{0.2}{0.6}\right)^2=2 \) mA, so options (c) and (d) are wrong.

Teacher's Note:
1. Stopping potential is independent of distance; saturation current varies as 1/d^2.

Answer: (d) a continuous spectrum with minimum wavelength about 0.155 Å and the characteristic X-ray spectrum of tungsten

Explanation:
1. \( \lambda_{min}=\frac{12.4}{80}=0.155 \) Å for the continuous spectrum.
2. Since 80 keV exceeds 72.5 keV, K electrons can be ejected, so characteristic lines also appear.

Teacher's Note:
1. Both spectra appear whenever the electron energy exceeds the K binding energy.

Answer: (a) \( \lambda_k-\lambda_c \) increases

Explanation:
1. \( \lambda_c=hc/eV \) decreases with V.
2. \( \lambda_k \) depends only on the target, so it is unchanged.
3. Therefore \( \lambda_k-\lambda_c \) increases.

Teacher's Note:
1. Characteristic lines do not move with voltage.

Answer: (a) \( \frac{2mc\lambda^2}{h} \)

Explanation:
1. For the electrons \( eV=K=\frac{h^2}{2m\lambda^2} \).
2. The cut-off wavelength is \( \lambda_0=\frac{hc}{eV}=\frac{hc\cdot2m\lambda^2}{h^2}=\frac{2mc\lambda^2}{h} \).

Teacher's Note:
1. Express the kinetic energy through the de Broglie wavelength first.

Answer: (c) only the ultraviolet lamps (50 W or 1 W ultraviolet)

Explanation:
1. Emission needs wavelength shorter than 5200 Å. Infrared has longer wavelength.
2. Ultraviolet is shorter, so ultraviolet lamps of any power cause emission, while no infrared lamp does.

Teacher's Note:
1. Frequency, not power, decides whether emission occurs.

Answer: (b) A suitable target needs a high melting point; the minimum wavelength is about \( 0.25\times10^{-10} \) m

Explanation:
1. Power is 1000 W, so 990 W heats the target: \( \frac{990}{95}\approx10.4 \) degrees C per second, not 2.
2. The target must withstand heating, so it needs a high melting point and high conductivity.
3. \( \lambda_{min}=\frac{12.4}{50}=0.248 \) Å, i.e. about \( 0.25\times10^{-10} \) m.

Teacher's Note:
1. Tungsten is used for its high melting point.

Answer: (d) A is false, but R is true.

Explanation:
1. Intensity is energy per unit time; a source of higher frequency emits fewer photons for the same power, so A is false.
2. R is therefore true.

Teacher's Note:
1. Number of photons per second = power / (hf).

Answer: (d) A is false, but R is true.

Explanation:
1. The metal lattice takes up the extra momentum, so momentum is conserved and A is false.
2. R is true.

Teacher's Note:
1. The rest of the metal absorbs the recoil momentum.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Maximum kinetic energy is \( hf-\phi \), independent of intensity, so A is true.
2. R is also true, but threshold frequency does not explain why K is independent of intensity.

Teacher's Note:
1. Each photon interacts with one electron.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A photon carries momentum \( p=E/c \) as a particle does, so R is true and supports A.
2. Both are true and R is the explanation.

Teacher's Note:
1. Having momentum is a particle property.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A grating works only when the spacing is comparable to the wavelength; X-ray wavelengths are about 1 Å, far smaller than the grating spacing.
2. So both are true and R explains A.

Teacher's Note:
1. Crystals serve as natural gratings for X-rays.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Electrons from different depths lose different energies before escaping, giving a spread: A is true.
2. The work function is a property of the metal (true) but does not explain the spread.

Teacher's Note:
1. Only the surface electrons get the maximum energy.

Answer: (c) A is true, but R is false.

Explanation:
1. All electromagnetic waves travel at c in vacuum, so A is true.
2. In glass the refractive index depends on frequency, so speeds differ: R is false.

Teacher's Note:
1. Dispersion arises from frequency-dependent refractive index.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. K lines come from transitions ending in n = 1, which have larger energy gaps, so shorter wavelength: R is true.
2. The shorter wavelength is not the reason for the emission mechanism, so R does not explain A.

Teacher's Note:
1. K lines are harder X-rays than L lines.

Answer: (d) A is false, but R is true.

Explanation:
1. Continuous X-rays do not depend on the target element, so they cannot identify it: A is false.
2. Characteristic lines follow Moseley's law and identify the element: R is true.

Teacher's Note:
1. Moseley used characteristic lines to order the elements.

Answer: (c) A is true, but R is false.

Explanation:
1. For a given line series, both a and b are constants (for example b = 1 for K-alpha), the same for all elements: A is true.
2. R says b depends on the material, which is false.

Teacher's Note:
1. The screening constant depends on the series, not on the element.

Answer: (a) 17.9 pm

Explanation:
1. \( \lambda_{min}=\frac{1240\ \text{eV nm}}{69500\ \text{eV}}=0.01784 \) nm.
2. This is about 17.9 pm.

Teacher's Note:
1. 1 pm = 0.001 nm.

Answer: (a) \( a^{-2} \)

Explanation:
1. Nodes at both ends require \( a=n\frac\lambda2 \), so \( \lambda=\frac{2a}{n} \).
2. Then \( E=\frac{p^2}{2m}=\frac{h^2}{2m\lambda^2}=\frac{n^2h^2}{8ma^2} \), proportional to \( a^{-2} \).

Teacher's Note:
1. Fit an integer number of half wavelengths into the box.

Answer: (d) n

Explanation:
1. Momentum \( p=\frac{h}{\lambda}=\frac{nh}{2a} \).
2. So \( v=p/m\propto n \).

Teacher's Note:
1. Wavelength is 2a/n, so momentum grows linearly with n.

Answer: (a) A-p, B-s, C-r, D-q

Explanation:
1. Thermal energy is about kT = 0.025 eV, so A-p.
2. Binding energy per nucleon is about 7 MeV (B-s), an X-ray photon is about a keV (C-r) and visible photons are about 2 eV (D-q).

Teacher's Note:
1. Memorise the order of magnitude of these energies.