Question

Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has a speed $v$. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be

• A. $>v^2{\left(\frac{4}{3}\right)}^{1/2}$
• B. $<v^2{\left(\frac{4}{3}\right)}^{1/2}$
• C. $=v{\left(\frac{4}{3}\right)}^{1/2}$
• D. $=v{\left(\frac{3}{4}\right)}^{1/2}$

Answer 1

The correct option is A $>v^2{\left(\frac{4}{3}\right)}^{1/2}$

$\begin{array}{c}\frac{nc}{\lambda }-\varnothing =\frac{1}{2}m{v}^2\to \left(1\right)\\ \frac{4hc}{3\lambda }-\varnothing =\frac{1}{2}m{v}_1^2\to \left(2\right)\\ \left(2\right)-\left(1\right)\\ \frac{4hc}{3\lambda }-\frac{nc}{\lambda }=\frac{1}{2}m\left(v_1^2-v^2\right)\\ \frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}^2+\frac{nc}{3\lambda }\\ v_1=\sqrt{v^2+\frac{2hc}{3\lambda m}}\to \left(3\right)\\ from\left(i\right)\frac{nc}{\lambda }=\varnothing +\frac{m{v}^2}{2}\\ \frac{3nc}{\lambda m}=\frac{2\varnothing }{m}{v}^2\\ \left(3\right)\&\left(4\right)\\ \frac{2\lambda c}{3\lambda m}=\frac{2\varnothing }{3m}+\frac{r^2}{3}>\frac{v^2}{3}\to \left(4\right)\\ v^2>v\sqrt{\frac{4}{3}}\end{array}$