JEE Advanced 2022 Paper 1 · Chemistry · Atomic Structure · medium

Photons emitted when He^+ ions undergo the transition n=4→ n=2 fall on a metal surface whose work function is 2.0 eV. Use En=-13.6 Z^2/n^2 eV…

Numerical answer

Photons emitted when

\mathrm{He^+}

ions undergo the transition

n=4\to n=2

fall on a metal surface whose work function is

2.0\ \mathrm{eV}

. Use

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

h=6.626\times10^{-34}\ \mathrm{J\,s}

m_e=9.109\times10^{-31}\ \mathrm{kg}

and

1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}

. The de Broglie wavelength of the fastest photoelectrons, in

\mathrm{nm}

, is (give the answer correct to two decimal places)

Show answer and solution

Answer: 0.43

For

\mathrm{He^+}

Z=2

. The photon energy is

E=13.6\times4\left(\dfrac1{2^2}-\dfrac1{4^2}\right)=54.4\times0.1875=10.2\ \mathrm{eV}

. By the photoelectric equation the maximum kinetic energy is

K=h\nu-\phi=10.2-2.0=8.2\ \mathrm{eV}=8.2\times1.602\times10^{-19}=1.314\times10^{-18}\ \mathrm{J}

. The momentum is

p=\sqrt{2mK}=\sqrt{2.394\times10^{-48}}=1.547\times10^{-24}\ \mathrm{kg\,m\,s^{-1}}

. Then

\lambda=h/p=6.626\times10^{-34}/1.547\times10^{-24}=4.283\times10^{-10}\ \mathrm{m}=0.43\ \mathrm{nm}

. Cross-check:

\lambda(\mathrm{nm})=1.226/\sqrt{K(\mathrm{eV})}=1.226/2.864=0.428

. Answer: 0.43.

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