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The radiation emitted, when an electron jumps from $n=3$ to $n=2$ orbit is a hydrogen atom, falls on a metal to produce photoelectron. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of $\frac{1}{320} \mathrm{~T}$ in a radius of $10^{-3} \mathrm{~m}$ . Find the work function of metal-
Option: 1
1.03 eV

Option: 2
1.89 Ev

Option: 3
0.86 eV

Option: 4
2.03 eV

Answers (1)

best_answer

$E_3-E_2=13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$

$=\frac{13.6 \times 5}{36}=1.89 \mathrm{eV}$

Photoelectrons with $\mathrm{KE}_{\max }$ are moving on circular path.

$r=\frac{m v}{q B}$

$\mathrm{mv}=\mathrm{qBr}$

$P=q B r=1.6 \times 10^{-19} \times \frac{1}{3200} \times 10^{-3}$

$\frac{1}{2} \times 10^{-24}=5 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

$\text { Energy of photoelectron }=\mathrm{KE}_{\max }=\frac{\mathrm{P}^2}{2 \mathrm{~m}}$

$=\frac{25 \times 10^{-50}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV}$

$=0.86 \mathrm{eV}$

Now use Einstein equation

$\mathrm{hv}=\phi+\mathrm{k} E_{\max }$

$0.56+\phi ; \quad \phi=1.03 \mathrm{ev}$

Posted by

Divya Prakash Singh

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