Question

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

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

Answer 1

Answer: (A)

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

Answer 2

$\frac{hc}{\lambda} -\phi = \frac{1}{2}mv^{2}$ .....(i) $\frac{4hc}{3\lambda} - \phi= \frac{1}{2} mv'^{2}$ ....(ii) $\frac{hc}{3\lambda} = \frac{1}{2}m \left(v'^{2} - v^{2}\right)$ $\Rightarrow v' = \sqrt{v^{2} + \frac{2hc}{3\lambda m}}$ ....(iii) also from $\frac{hc}{\lambda } = \phi + \frac{mv^{2}}{2}$ $\Rightarrow \frac{2hc}{\lambda m} = \frac{2\phi}{m} + v^{2}$ /. $\Rightarrow \frac{2hc}{3\lambda m} = \frac{2\phi}{3m} + \frac{r^{2}}{3} > \frac{v^{2}}{3}$ .....(iv) combining (iii) & (iv) $v' > \sqrt{\frac{4v^{2}}{3}}$