Question

The photoelectric threshold wavelength of silver is$3250\times {10}^{-10}$ m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536\times {10}^{-10}m$ is (Given $h=4.14\times$${10}^{-15}$ eVs and $c=3\times {10}^{8}m{s}^{-1})$:

• A. $6\times {10}^5{ms}^{-1}$
• B. $0.6\times {10}^6{ms}^{-1}$
• C. $61\times {10}^3{ms}^{-1}$
• D. $0.3\times {10}^6{ms}^{-1}$

Answer 1

The correct option is A $6\times {10}^5{ms}^{-1}$

$\begin{array}{c}{\lambda }_0=3250\times {10}^{-10}m=3250A^0\\ \lambda =2536\times {10}^{-10}m=2536A^0\\ \frac{1}{2}m{v}^2=hc\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]\\ v=\sqrt{\frac{2hc}{m}\left[\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right]}\\ =\sqrt{\frac{2\times 12400\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\left[\frac{714}{2536\times 3250}\right]}\\ =0.6\times {10}^{6}m/s=6\times {10}^{5}m/s\end{array}$

Hence,

option $\left(A\right)$ is correct answer.