Question

The radiation corresponding to $3\to 2$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3\times {10}^{-4}\text{T}$. If the radius of the largest circular path followed by these electrons is $10.0\text{mm}$, the work function of the metal is close to

• A. $1.6\text{eV}$
• B. $1.1\text{eV}$
• C. $1.8\text{eV}$
• D. $0.8\text{eV}$

Answer 1

The correct option is B $1.1\text{eV}$

When an electron moves in a circular path, then
$evB=\frac{m{v}^2}{r}r=\frac{mv}{eB}r=\frac{\sqrt{2meV}}{eB}\text{or}\frac{1}{B}\sqrt{\frac{2mV}{e}}V=\frac{e{r}^2{B}^2}{2m}\to \text{stopping potential}$
$V=\frac{100\times {10}^{-6}\times 1.6\times {10}^{-19}\times 9\times {10}^{-8}}{2\times 9.1\times {10}^{-31}}=0.8V$

KE of electron, $K=0.8\text{eV}$

For transition between $3$ to $2$,

$E=13.6(\frac{1}{2^2}-\frac{1}{3^2})$
$E=\frac{13.6\times 5}{36}E=1.88\text{eV}$

Work fuction,
${\varphi }_o=E-K_{max}$
$=1.89-0.79\approx 1.1\text{eV}$

Hence, option $B$ is correct.