JEE Main 2019 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

When light of wavelength 400 nm falls on a metal surface, the stopping potential is 1.0 V. If the surface is illuminated with light of wavelength 300…

Single correct option

When light of wavelength

400\ \mathrm{nm}

falls on a metal surface, the stopping potential is

1.0\ \mathrm{V}

. If the surface is illuminated with light of wavelength

300\ \mathrm{nm}

instead, the maximum kinetic energy of the emitted photoelectrons is nearly (use

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

)

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Answer: (A) $2.03\ \mathrm{eV}$

First find the work function from the

400\ \mathrm{nm}

data: photon energy

=1240/400=3.1\ \mathrm{eV}

, so

\phi = 3.1 - 1.0 = 2.1\ \mathrm{eV}

(the maximum kinetic energy equals

eV_0=1.0\ \mathrm{eV}

). At

300\ \mathrm{nm}

the photon energy is

1240/300\approx4.13\ \mathrm{eV}

, hence

K_{max}=4.13-2.1\approx2.03\ \mathrm{eV}

. The threshold wavelength is

1240/2.1\approx590\ \mathrm{nm}

, which is longer than both wavelengths, so emission occurs in both cases.

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