A He^+ ion (hydrogen-like, Z=2) makes…

A He^+ ion (hydrogen-like, Z=2) makes a transition from the level n=4 to n=2 and the emitted photon falls on a metal surface of work function 4.2 eV.…

Single correct option

\mathrm{He^+}

ion (hydrogen-like,

Z=2

) makes a transition from the level

n=4

n=2

and the emitted photon falls on a metal surface of work function

4.2\ \mathrm{eV}

. The photoelectron of maximum kinetic energy then moves freely, with no field acting on it. 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}

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

. The de Broglie wavelength of this photoelectron is closest to

A. $0.38\ \mathrm{nm}$
B. $0.71\ \mathrm{nm}$
C. $0.50\ \mathrm{nm}$
D. $0.43\ \mathrm{nm}$

Show answer and solution

Answer: (C) $0.50\ \mathrm{nm}$

Photon energy:

E=13.6\times2^2\left(\dfrac1{2^2}-\dfrac1{4^2}\right)=54.4\times0.1875=10.2\ \mathrm{eV}

.
Maximum kinetic energy of the photoelectron:

K=10.2-4.2=6.0\ \mathrm{eV}=9.612\times10^{-19}\ \mathrm{J}

.
de Broglie wavelength:

\lambda=\dfrac{h}{\sqrt{2m_eK}}=\dfrac{6.626\times10^{-34}}{\sqrt{2\times9.109\times10^{-31}\times9.612\times10^{-19}}}=5.0\times10^{-10}\ \mathrm{m}=0.50\ \mathrm{nm}

.
(A)

0.38\ \mathrm{nm}

results from using the whole photon energy

10.2\ \mathrm{eV}

as kinetic energy (forgetting the work function); (B)

0.71\ \mathrm{nm}

results from a kinetic energy of

3.0\ \mathrm{eV}

(half of the correct value); (D)

0.43\ \mathrm{nm}

corresponds to

8.1\ \mathrm{eV}

(subtracting

2.1\ \mathrm{eV}

). Answer: (C).

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Original question written in the JEE Advanced 2023 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake