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. $1.03eV$
• B. $1.89eV$
• C. $0.86eV$
• D. $2.03eV$

Answer 1

Answer: (A)

$1.03eV$
$E_3-E_2=13.6\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{13.6\times 5}{36}=1.89eV$
Photoelectrons with $k{E}_{max}$ are moving on circular path.
$r=\frac{mv}{qB}$
$\Rightarrow mv=qBr$
$\Rightarrow P=qBr$
$\Rightarrow P=1.6\times 10^{-19}\times \frac{1}{3200}\times 10^{-3}$
$\Rightarrow \frac{1}{2}\times 10^{-24}=5\times 10^{-25}kgm/s$
Energy of photoelectron $\Rightarrow K{E}_{max}=\frac{P^2}{2m}$
$\Rightarrow K{E}_{max}=\frac{25\times 10^{-50}}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}}eV$
$\Rightarrow K{E}_{max}=0.86eV$
Now use Einstein equation
$hv=\varphi +K{E}_{max}$
$\Rightarrow 1.89=0.56+\varphi ;$
$\Rightarrow \varphi =1.03eV$