Question

If light of wavelength ${\lambda }_1$ is allowed to fall on a metal, then kinetic energy of photoelectrons emitted is $K_1$. If wavelength of light changes to ${\lambda }_2$, then kinetic energy of electron changes to $K_2$. Then, work function of the metal is

• A. $\frac{K_1{K}_2({\lambda }_1-{\lambda }_2)}{{\lambda }_1{\lambda }_2}$
• B. $\frac{K_1{\lambda }_1-K_2{\lambda }_2}{({\lambda }_1-{\lambda }_2)}$
• C. $\frac{K_1{\lambda }_1-K_2{\lambda }_2}{({\lambda }_2-{\lambda }_1)}$
• D. $\frac{{\lambda }_1{\lambda }_2{K}_1{K}_2}{({\lambda }_2-{\lambda }_1)}$

Answer 1

The correct option is C $\frac{K_1{\lambda }_1-K_2{\lambda }_2}{({\lambda }_2-{\lambda }_1)}$

We know the equation for kinetic energy in photoelectric effect as
$\frac{hc}{\lambda }=K+\varphi$
where $K$ is the kinetic energy of the electron coming out, $\varphi$ is the work function of the material and $\lambda$ is the wavelength of the photon.
Now for first case
$\frac{hc}{{\lambda }_1}=K_1+\varphi$ .....(i)
$\frac{hc}{{\lambda }_2}=K_2+\varphi$ .....(ii)
eq(i)/eq(ii)
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{K_1+\varphi }{K_2+\varphi }$
${\lambda }_2{K}_2+{\lambda }_2\varphi ={\lambda }_1{K}_1+{\lambda }_1\varphi$
On rearranging
$\varphi =\frac{K_1{\lambda }_1-K_2{\lambda }_2}{{\lambda }_2-{\lambda }_1}$