JEE Physics Dual Nature Of Radiation And Matter MCQs Set 13

Answer: (b) \( \sqrt{\frac{2eV}{m}} \)

Explanation:
1. Work done equals kinetic energy gained: \( eV=\frac12mv^2 \).
2. Hence \( v=\sqrt{2eV/m} \).

Teacher's Note:
1. Energy gained = charge x potential difference.
2. The same relation gives the de Broglie wavelength \( \lambda=h/\sqrt{2meV} \).

Answer: (b) \( \lambda\sqrt{\frac mM} \)

Explanation:
1. For a particle of charge e: \( \lambda=\frac{h}{\sqrt{2meV}} \), so \( \lambda\propto\frac1{\sqrt m} \).
2. Therefore \( \lambda_p=\lambda\sqrt{m/M} \).

Teacher's Note:
1. Same charge and potential means same kinetic energy; the heavier particle has the smaller wavelength.

Answer: (a) 2 : 1

Explanation:
1. At equal kinetic energy, \( \lambda=\frac{h}{\sqrt{2mK}}\propto\frac1{\sqrt m} \).
2. \( \frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_p}}=\sqrt4=2 \).

Teacher's Note:
1. Compare at equal K using lambda = h / sqrt(2mK).
2. For equal momentum the wavelengths would be equal.

Answer: (c) 1

Explanation:
1. Initial momentum is zero, so the products have equal and opposite momenta: \( p_1=p_2 \).
2. As \( \lambda=h/p \), the wavelengths are equal and the ratio is 1.

Teacher's Note:
1. Momentum conservation fixes equal magnitudes of momentum after a decay from rest.

Answer: (d) the intensity of the incident light increases

Explanation:
1. Each photon above the threshold frees at most one electron, so the number of electrons per second is proportional to the number of photons per second.
2. The number of photons per second is set by the intensity of the light.

Teacher's Note:
1. Intensity controls the photocurrent; frequency controls the maximum kinetic energy.

Answer: (a) the photoelectric current increases

Explanation:
1. More photons per second eject more electrons per second, so the current increases.
2. The maximum kinetic energy \( hf-\phi \) depends only on the frequency, so it is unchanged.

Teacher's Note:
1. Intensity changes how many electrons come out, not how fast they come out.

Answer: (b) \( 3.2\times10^{-19} \) J

Explanation:
1. \( K_{max}=6.2-4.2=2.0 \) eV.
2. Converting, \( 2.0\times1.6\times10^{-19}=3.2\times10^{-19} \) J.

Teacher's Note:
1. Always convert eV to joule by multiplying by 1.6 x 10^-19.

Answer: (b) 4

Explanation:
1. At the stopping potential \( eV_0=K_{max} \).
2. So \( V_0=4 \) V.

Teacher's Note:
1. The stopping potential in volts equals the maximum kinetic energy in eV.

Answer: (d) 1.4 V

Explanation:
1. \( eV_0=h f-\phi=2-0.6=1.4 \) eV.
2. So the stopping potential is 1.4 V.

Teacher's Note:
1. Stopping potential = (photon energy - work function) / e.

Answer: (a) no electrons are emitted

Explanation:
1. Green light has higher frequency than yellow, so the threshold frequency lies between yellow and green.
2. Red has lower frequency than yellow, hence below threshold, and no electron is emitted.

Teacher's Note:
1. Below the threshold frequency there is no emission however intense the light.

Answer: (b) 1 : 2

Explanation:
1. Let \( \phi_A=\phi \), \( \phi_B=2\phi \). Then \( K_A=hf-\phi \) and \( K_B=2hf-2\phi=2(hf-\phi) \).
2. Hence \( K_A:K_B=1:2 \).

Teacher's Note:
1. Both the photon energy and the work function double for B.

Answer: (b) 1 : 4

Explanation:
1. \( K_1=1-0.5=0.5 \) eV and \( K_2=2.5-0.5=2.0 \) eV.
2. Ratio \( 0.5:2.0=1:4 \).

Teacher's Note:
1. Apply K = E - phi for each photon energy separately.

Answer: (a) \( I_1=I_2 \)

Explanation:
1. The photocurrent depends on the number of photons striking the cathode per second, which is unchanged.
2. Because every photon is still above threshold, the same number of electrons is emitted; only their kinetic energy is lower.

Teacher's Note:
1. Work function changes the kinetic energy, not the saturation current.

Answer: (b) \( v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2) \)

Explanation:
1. For the same work function: \( hf_1-\phi=\frac12mv_1^2 \) and \( hf_2-\phi=\frac12mv_2^2 \).
2. Subtracting: \( \frac12m(v_1^2-v_2^2)=h(f_1-f_2) \), i.e. \( v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2) \).

Teacher's Note:
1. Subtracting the Einstein equations removes the common work function.

Answer: (a) X-rays

Explanation:
1. Failure with ultraviolet means its frequency is below threshold.
2. Only radiation of higher frequency can work, and X-rays have a higher frequency than ultraviolet.

Teacher's Note:
1. Order by frequency: radio, infrared, visible, ultraviolet, X-rays.

Answer: (b) voltage applied to the tube

Explanation:
1. The most energetic photon carries the whole kinetic energy of an electron: \( eV=\frac{hc}{\lambda_{min}} \).
2. So \( \lambda_{min}=\frac{12400}{V} \) Å depends only on the tube voltage (Duane-Hunt law).

Teacher's Note:
1. The cut-off wavelength is independent of target material and tube current.

Answer: (c) \( f_2=f_1+f_3 \)

Explanation:
1. \( K_\alpha \) is the L to K transition, \( K_\beta \) the M to K, and \( L_\alpha \) the M to L transition.
2. Since M to K = (M to L) + (L to K), energies add: \( f_2=f_1+f_3 \).

Teacher's Note:
1. Draw the K, L, M levels and add the energy gaps.

Answer: (c) n = 3 to n = 1

Explanation:
1. Transitions ending in the K shell (n = 1) give the K series.
2. \( K_\alpha \) is 2 to 1 and \( K_\beta \) is 3 to 1.

Teacher's Note:
1. Greek subscripts count how many shells the electron falls from.

Answer: (d) 40 kV

Explanation:
1. \( V=\frac{12400}{\lambda_{min}(\text{Å})}=\frac{12400}{0.3094} \).
2. This is about 40 000 V = 40 kV.

Teacher's Note:
1. Use lambda_min (in angstrom) = 12400 / V (in volt).

Answer: (c) All wavelengths in it are larger than a certain minimum wavelength

Explanation:
1. The continuous spectrum has a sharp short-wavelength limit \( \lambda_{min}=hc/eV \).
2. There is no sharp upper limit; hence all wavelengths exceed the minimum value.

Teacher's Note:
1. The beam is a continuous spectrum with a few characteristic lines superimposed.

Answer: (c) \( V>40 \) kV

Explanation:
1. Characteristic X-rays arise when an incident electron ejects an inner (K) electron.
2. The incident electron must carry more than 40 keV, so V must be more than 40 kV.

Teacher's Note:
1. Energy must exceed the binding energy to eject the electron.

Answer: (d) 0.31 Å

Explanation:
1. \( \lambda_{min}=\frac{hc}{E}=\frac{12.4\ \text{keV Å}}{40\ \text{keV}} \).
2. This gives \( \lambda_{min}\approx0.31 \) Å.

Teacher's Note:
1. The photon with the highest energy has the smallest wavelength.

Answer: (c) 59 keV

Explanation:
1. \( \Delta E=\frac{hc}{\lambda}=\frac{1240\ \text{eV nm}}{0.021\ \text{nm}} \).
2. This is about \( 5.9\times10^4 \) eV = 59 keV.

Teacher's Note:
1. K-alpha photon energy equals the K-L gap.

Answer: (d) \( \lambda\propto\frac1{(Z-1)^2} \)

Explanation:
1. Moseley's law gives \( \sqrt f=a(Z-b) \) with \( b=1 \) for the K-alpha line.
2. So \( f\propto(Z-1)^2 \) and \( \lambda\propto\frac1{(Z-1)^2} \).

Teacher's Note:
1. The screening constant for K-alpha is 1.

Answer: (a) is the same for all metals and independent of the intensity of radiation

Explanation:
1. \( K_{max}=hf-\phi \) is a straight line with slope h.
2. h is a universal constant, so the slope is the same for all metals and for every intensity.

Teacher's Note:
1. Different metals only shift the intercept on the frequency axis.