Question

In a photo-emissive cell, with exciting wavelength $\lambda$, the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $3 \lambda / 4$, the kinetic energy of the fastest emitted electron will be

• A. $\frac{3 K}{4}$
• B. $\frac{4 K}{3}$
• C. less than $\frac{4 K}{3}$
• D. more than $\frac{4 K}{3}$

Answer 1

Answer: (D)

$K=\frac{h c}{\lambda}-\phi_{0}$ ...(i)
and $K'=\frac{4 h c}{3 \lambda}-\phi_{0}$ ...(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow K'-K=\frac{4 h c}{3 \lambda}-\frac{h c}{\lambda}$
$K'-K=\frac{h c}{3 \lambda}$
But from Eq. (i) $\frac{h c}{\lambda}=K+\phi_{0}$
$\therefore K'-K=\frac{K}{3}+\frac{\phi_{0}}{3}$
$\Rightarrow K'=\frac{4 K}{3}+\frac{\phi_{0}}{3}$
Or $K' > \frac{4 K}{3}$