Question

When the wavelength of radiation falling on a metal is changed from $500\text{nm}$ to $200\text{nm}$ the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to-

• A. $\text{0.81 eV}$
• B. $\text{1.02 eV}$
• C. $\text{0.52 eV}$
• D. $\text{0.61 eV}$

Answer 1

The correct option is D $\text{0.61 eV}$

Using Einstein's photoelectric equation,

$\frac{hc}{\lambda }=\varphi +K{E}_{max}$

$K{E}_{max}=\frac{hc}{\lambda }-\varphi =\frac{hc}{500}-\varphi ....\left(1\right)$

$3K{E}_{max}=\frac{hc}{\lambda }-\varphi =\frac{hc}{200}-\varphi ....\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$,

$\frac{3}{1}=\frac{\frac{hc}{200}-\varphi }{\frac{hc}{500}-\varphi }$

$\therefore \varphi =\frac{hc}{2000}=0.61\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.