Question

The photoelectric threshold wavelength of silver is $3250\times {10}^{-10}m$. The velocity of the electron ejected from a silver surface by ultravioet light of wavelength $2536\times {10}^{-10}m$ is:

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

Answer 1

Answer: (C)

. $\approx 0.3\times {10}^{6}m{s}^{-1}$.

Given,

Threshold wavelength = ${\lambda }_0=3250\times {10}^{-10}m$

Incident wavelength = $\lambda =2536\times {10}^{-10}m$

From the Einstein’s Photoelectric Equation we have,

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

Therefor maximum kinetic energy $K{E}_{max}=\frac{h_c}{\lambda }-\varphi$

$K{E}_{max}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})=\frac{1}{2}{mv}^2$

$v=\sqrt{\frac{{2h}_c}{m}(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})}$

$h=6.67\times {10}^{-34}J.s\text{and}c=3\times {10}^{8}m/s$

Putting all these values we get,

$v=0.3\times {10}^{6}m/s$.