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

• A. $1.03 \, eV$
• B. $1.89 \, eV$
• C. $0.86 \, eV$
• D. $2.0 \, 3 \, eV$

Answer 1

Answer: (A)

$1.03 \, eV$
$E_{3}-E_{2}=13.6\left[\frac{1}{2^{2}} - \frac{1}{3^{2}}\right]=\frac{13.6 \times 5}{36}=1.89 \, eV$
Photoelectrons with $kE_{m a x}$ are moving on circular path.
$r=\frac{m v}{q B}$
$\Rightarrow mv=qBr$
$\Rightarrow P=qBr$
$\Rightarrow P=1.6\times 10^{- 19}\times \frac{1}{3200}\times 10^{- 3}$
$\Rightarrow \frac{1}{2}\times 10^{- 24}=5\times 10^{- 25}kg m / s$
Energy of photoelectron $\Rightarrow KE_{m a x}=\frac{P^{2}}{2 \, m}$
$\Rightarrow K E_{m a x}=\frac{25 \times 10^{- 50}}{2 \times 9.1 \times 1 0^{- 31} \times 1.6 \times 1 0^{- 19}}eV$
$\Rightarrow K E_{m a x}=0.86 \, eV$
Now use Einstein equation
$hv=\phi+KE_{m a x}$
$\Rightarrow 1.89=0.56+\phi;$
$\Rightarrow \phi=1.03 \, eV$