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

Answer 1

Answer: (D)

$K=\frac{hc}{\lambda }-{\varphi }_0$ ...(i)
and $K^{\prime }=\frac{4hc}{3\lambda }-{\varphi }_0$ ...(ii)
From Eqs. (i) and (ii), we get
$\Rightarrow K^{\prime }-K=\frac{4hc}{3\lambda }-\frac{hc}{\lambda }$
$K^{\prime }-K=\frac{hc}{3\lambda }$
But from Eq. (i) $\frac{hc}{\lambda }=K+{\varphi }_0$
$\therefore K^{\prime }-K=\frac{K}{3}+\frac{{\varphi }_0}{3}$
$\Rightarrow K^{\prime }=\frac{4K}{3}+\frac{{\varphi }_0}{3}$
Or $K^{\prime }>\frac{4K}{3}$