Question

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

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

Answer 1

Answer: (D)

Greater then $v(4/3)^{1/2}$

Answer 2

From $E = W_{0} + \frac{1}{2}mv_{\max}^{2}$

⇒ ${v\sqrt{\frac{2E}{m} - \frac{2W_{0}}{m}}}_{\max}$ (where $E = \frac{hc}{\lambda}$)

If wavelength of incident light charges from λ 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 $v' = \sqrt{\frac{2E'}{m} - \frac{2W_{0}}{m}}$ …….(ii)

As $E \propto \frac{1}{\lambda}$

⇒ $E' = \frac{4}{3}E$ hence $v' = \sqrt{\frac{2\left( \frac{4}{3}E \right)}{m} - \frac{2W_{0}}{m}}$

⇒ $v' = \left( \frac{4}{3} \right)^{1/2}\sqrt{\frac{2E}{m} - \frac{2W_{0}}{m\left( \frac{4}{3} \right)^{1/2}}}$

⇒ $v' = \left( \frac{4}{3} \right)^{1/2}$ $X = \sqrt{\frac{2E}{m} - \frac{2W_{0}}{m\left( \frac{4}{3} \right)^{1/2}}} > v$

so $v' > \left( \frac{4}{3} \right)^{1/2}v$.