Question

Radiation, with wavelength $6561\mathring{\text{A}}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3\times {10}^{-4}\text{T}$. If the radius of the largest circular path followed by the electrons is $10\text{mm}$, the work function of the metal is close to:

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

Answer 1

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

Using Einstein's photoelectric equation,
$E={\varphi }_0+(KE{)}_{max}$

${\varphi }_0=\frac{hc}{\lambda }-(KE{)}_{max}=\frac{12420}{\lambda }-(KE{)}_{max}\left(\text{in eV}\right)$

Also, $(KE{)}_{max}=\frac{p^2}{2m}$

For an electron moving in a circular path, through a uniform magnetic field,

$r=\frac{mv}{Bq}=\frac{p}{eB}\Rightarrow p=reB$

$\Rightarrow (KE{)}_{max}=\frac{r^2{e}^2{B}^2}{2m}\text{in V}=\frac{r^{2}e{B}^2}{2m}(\text{in eV})$

Now, ${\varphi }_0=\frac{12420}{\lambda }-\left(KE{)}_{max}\right(\text{in eV})$

$\Rightarrow {\varphi }_0=\frac{12420}{6561}-\frac{r^{2}e{B}^2}{2m}\left(\text{in eV}\right)$

$=1.89-\frac{\left({10}^{-4}\right)\times 1.6\times {10}^{-19}\times 9\times {10}^{-8}}{2\times 9.07\times {10}^{-31}}$

$=(1.893-0.793)=1.1\text{eV}$

Hence, $\left(A\right)$ is the correct answer.