A photon emitted when the electron of…

A photon emitted when the electron of a Li^2+ ion falls from the n=3 level to the n=2 level falls on a hydrogen atom in its ground state and ejects…

Single correct option

A photon emitted when the electron of a

\mathrm{Li^{2+}}

ion falls from the

n=3

level to the

n=2

level falls on a hydrogen atom in its ground state and ejects its electron. Taking the Bohr energy

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 ejected electron is closest to

Show answer and solution

Answer: (B) $0.665\ \mathrm{nm}$

Photon energy from

\mathrm{Li^{2+}}

(

Z=3

E=13.6\times9\left(\dfrac14-\dfrac19\right)=17.0\ \mathrm{eV}

. The hydrogen atom in its ground state has ionisation energy

13.6\ \mathrm{eV}

, so the kinetic energy of the photoelectron is

K=17.0-13.6=3.4\ \mathrm{eV}=5.447\times10^{-19}\ \mathrm{J}

. The de Broglie wavelength is

\lambda=\dfrac{h}{\sqrt{2m_eK}}=\dfrac{6.626\times10^{-34}}{\sqrt{2(9.109\times10^{-31})(5.447\times10^{-19})}}=6.65\times10^{-10}\ \mathrm{m}=0.665\ \mathrm{nm}

. If the whole photon energy

17\ \mathrm{eV}

were used as kinetic energy (forgetting the work done against ionisation) one would get

0.297\ \mathrm{nm}

; using

2K

K/2

gives

0.470

0.941\ \mathrm{nm}

. Hence

0.665\ \mathrm{nm}

Original question written in the JEE Advanced 2023 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake