JEE Main 2020 · Chemistry · Atomic Structure · medium

Photons emitted when the electron of Li^2+ falls from n=3 to n=2 are incident on a metal of work function 4.0 eV. Using the Bohr energy En=-13.6…

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Photons emitted when the electron of

\mathrm{Li^{2+}}

falls from

n=3

n=2

are incident on a metal of work function

4.0\ \mathrm{eV}

. Using the Bohr energy

E_n=-13.6\,Z^2/n^2\ \mathrm{eV}

, the maximum kinetic energy of the photoelectrons is

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Answer: (A) $13.0\ \mathrm{eV}$

Photon energy

=13.6\times3^2\left(\dfrac1{2^2}-\dfrac1{3^2}\right)=122.4\times\dfrac5{36}=17.0

eV. By the photoelectric equation,

K_{max}=h\nu-\phi=17.0-4.0=13.0

eV. Using

Z=1

would give a photon of only

1.89

eV, too small to eject electrons.

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