Question

If the light of wavelength $\lambda$ is incident on metal surface, the ejected fastest electron has speed $v$. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be

• A. Smaller than $\sqrt{\frac{4}{3}}v$
• B. Greater than $\sqrt{\frac{4}{3}}v$
• C. $2v$
• D. Zero

Answer 1

The correct option is B Greater than $\sqrt{\frac{4}{3}}v$

According to photoelectric law:

${KE}_{max}=\frac{1}{2}m{V}_{max}^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}\to V_{max}=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\lambda }{{\lambda }_0\lambda }\right)}-\left(1\right)\Rightarrow V_{max}^{\prime }=\sqrt{\frac{2hc}{m}\left(\frac{{\lambda }_0-\frac{3}{4}\lambda }{{\left(\frac{3}{4}\right)\lambda }_0\lambda }\right)}-\left(2\right)Divide\left(2\right)by\left(1\right):\frac{V_{max}^{\prime }}{V_{max}}=\sqrt{\frac{4}{3}}\Rightarrow V_{max}^{\prime }=\sqrt{\frac{4}{3}}{V}_{max}\to V^{\prime }=\sqrt{\frac{4}{3}}V$