Question

Light of wavelength $\lambda$ and photon energy 3eV falling on a metal surface produces photo electrons of maximum velocity V. If the wavelength of the incident light is changed to $\frac{3}{4}\lambda$ and the maximum velocity of the electron is doubled, the work function of the metal is nearly

• A. 1.66 eV
• B. 2.66 eV
• C. 3.66 eV
• D. 4.66 eV

Answer 1

Answer: (B)

$\begin{array}{c}E=hv-\varphi \\ \Rightarrow \frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi .........\left(1\right)\\ \Rightarrow \frac{1}{2}m{\left(2v\right)}^2=4\frac{hc}{3\lambda }-\varphi ..........\left(2\right)\\ divideequation\left(1\right)and\left(2\right)\\ \Rightarrow \frac{v^2}{4v^2}=\frac{\frac{hc}{\lambda }-\varphi }{\frac{4hc}{\lambda }-\varphi }\\ \Rightarrow \frac{4hc}{\lambda }-4\varphi =\frac{4hc}{3\lambda }-\varphi \\ \Rightarrow \frac{4hc}{\lambda }-\frac{4hc}{3\lambda }=3\varphi \\ \Rightarrow \frac{8hc}{3\lambda }=3\varphi \\ \Rightarrow \frac{hc}{\lambda }=3ev\left(given\right)\\ \Rightarrow \varphi =\left(\frac{8}{9}\times 3\right)ev\\ =2.66ev\end{array}$

Hence,

option $B$ is correct.