JEE Advanced 2018 Paper 2 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

Monochromatic light of wavelength 310 nm is incident on a metal surface of work function 2.5 eV. Use hc=1240 eV nm and the de Broglie relation λdB…

One or more options correct

Monochromatic light of wavelength

310\ \mathrm{nm}

is incident on a metal surface of work function

2.5\ \mathrm{eV}

. Use

hc=1240\ \mathrm{eV\,nm}

and the de Broglie relation

\lambda_{dB}\,[\mathrm{nm}]=1.226/\sqrt{K\,[\mathrm{eV}]}

for an electron of kinetic energy

K

. Which of the following statements is/are correct?

A. The stopping potential for this light is $4.0\ \mathrm{V}$
B. The threshold wavelength of the metal is $496\ \mathrm{nm}$
C. If the intensity of the light is doubled, the stopping potential also doubles
D. The shortest de Broglie wavelength among the emitted electrons is $0.50\ \mathrm{nm}$

Show answer and solution

Answer: B

The photon energy is

hc/\lambda=1240/310=4.0\ \mathrm{eV}

, so by Einstein's equation

K_{max}=4.0-2.5=1.5\ \mathrm{eV}

and the stopping potential is

1.5\ \mathrm{V}

, not

4.0\ \mathrm{V}

; (A) is wrong. The threshold wavelength is

\lambda_0=hc/\phi=1240/2.5=496\ \mathrm{nm}

, so (B) is correct. Doubling the intensity doubles the number of photons and hence the saturation current, but

K_{max}

, and therefore the stopping potential, depends only on frequency and work function and stays

1.5\ \mathrm{V}

; (C) is wrong. The fastest photoelectrons have the shortest de Broglie wavelength

\lambda=1.226/\sqrt{1.5}\approx1.0\ \mathrm{nm}

, not

0.50\ \mathrm{nm}

(that value would need

K\approx6\ \mathrm{eV}

); (D) is wrong. Answer: B only.

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