Question

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

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

Answer 1

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

Kinetic energy of emitted electron $E_1=\frac{hc}{\lambda }-\varphi$

where $\lambda$ is incident wavelength of radiation and $\varphi$ is the work function of the photocell.

$\therefore$ $\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi ..........\left(i\right)$

In second case : Incident wavelength is $\frac{3\lambda }{4}$.

Thus we get $E_2=\frac{4hc}{3\lambda }-\varphi$

Or $\frac{1}{2}m{v}_2^2=\frac{4hc}{3\lambda }-\varphi .........\left(ii\right)$

Clearly, $E_2>\frac{4}{3}{E}_1$ and hence $v_2>v{\left(\frac{4}{3}\right)}^{1/2}$.