Question

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 ${\displaystyle \frac{1}{320}}$T in a radius of 10^-3m. Find the 320 work function of metal:

Options

• 1.03 eV

• 1.89 eV

• 0.86 eV

• 2.03 eV

Answer 1

1.03 eV

Explanation:

E₃ – E₂ = 13.6 ${\displaystyle [\frac{1}{2^2}-\frac{1}{3^2}]}$

= ${\displaystyle \frac{13.6\times 5}{36}}$

= 1.89 eV

Photoelectrons with kE_max are moving on circular path.

r = ${\displaystyle \frac{\text{mv}}{\text{qB}}}$

mv = qBr

P = qBr

= ${\displaystyle 1.6\times {10}^{–19}\frac{1}{3200}\times {10}^{-3}}$

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

Energy of photoelectron = KE_max = ${\displaystyle \frac{{\text{P}}^2}{2\text{m}}}$

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

= 0.86 eV

Now use Einstein equation

hn = Φ + kE_max

1.89 = 0.56 + Φ;

Φ = 1.03 eV