Question

In a photoemissive cell with exciting wavelength $\lambda$ , the fastest electron has speed v. If the exciting wavelength is changed to $\frac{3\lambda }{4}$ , the speed of the fastest emitted electron will be

• A. $v{\left(\frac{3}{4}\right)}^{1/2}$
• B. $v{\left(\frac{4}{3}\right)}^{1/2}$
• C. Less than $v{\left(\frac{4}{3}\right)}^{1/2}$
• D. Greater then $v{\left(\frac{4}{3}\right)}^{1/2}$

Answer 1

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

From $E=W_0+\frac{1}{2}m{v}_{max}^2\Rightarrow v_{max}=\sqrt{\frac{2E}{m}-\frac{2W_0}{m}}$ (where $E=\frac{hc}{\lambda }$)
If wavelength of incident light charges from $\lambda$ to $\frac{3\lambda }{4}$ (decreases)
Let energy of incident light charges from E to E’ and speed of fastest electron changes from v to v’ then
$v=\sqrt{\frac{2E}{m}-\frac{2W_0}{m}}$ . . . . . . .(i) and $v^{\prime }=\sqrt{\frac{2E^{\prime }}{m}-\frac{2W_0}{m}}$ . . . . . .(2)
As $E\infty \frac{1}{\lambda }\Rightarrow E^{\prime }=\frac{4}{3}E$ hence $v^{\prime }=\sqrt{\frac{2\left(\frac{4}{3}E\right)}{m}-\frac{2W_0}{m}}\Rightarrow v^{\prime }={\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{2E}{m}-\frac{2W_0}{m{\left(\frac{4}{3}\right)}^{\frac{1}{2}}}}$
$\Rightarrow v^{\prime }={\left(\frac{4}{3}\right)}^{\frac{1}{2}}\sqrt{\frac{2E}{m}-\frac{2W_0}{m{\left(\frac{4}{3}\right)}^{\frac{1}{2}}}}$ > v