JEE Main 2016 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

For a certain metal the stopping potential is 2 V for light of frequency nu and 6 V for light of frequency 2nu. Taking hc=12400 eV AA, the threshold…

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For a certain metal the stopping potential is

2\ \mathrm{V}

for light of frequency

\nu

and

6\ \mathrm{V}

for light of frequency

2\nu

. Taking

hc=12400\ \mathrm{eV\,\AA}

, the threshold wavelength of the metal is

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Answer: (C) $6200\ \mathrm{\AA}$

Einstein's equation gives

h\nu=\phi+eV_0

. For the two frequencies:

h\nu=\phi+2

and

2h\nu=\phi+6

(energies in eV). Subtracting the first from the second:

h\nu=4\ \mathrm{eV}

, hence

\phi=4-2=2\ \mathrm{eV}

. The threshold wavelength is

\lambda_0=\dfrac{hc}{\phi}=\dfrac{12400}{2}=6200\ \mathrm{\AA}

. Check: the threshold frequency is

\nu_0=\phi/h=\nu/2

, so

\lambda_0=2\lambda

where

\lambda=\dfrac{12400}{4}=3100\ \mathrm{\AA}

is the wavelength of the first light.

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