Question

In a photoelectric experiment, with light of wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $5\lambda /4$, the speed of the fastest emitted electron will become

• A. $v\sqrt{\frac{5}{4}}$
• B. $v\sqrt{\frac{5}{3}}$
• C. less than $v\sqrt{\frac{5}{3}}$
• D. greater than $v\sqrt{\frac{5}{3}}$

Answer 1

Answer: (D)

$\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\varphi$
$\frac{1}{2}m{v}^{\prime 2}=\frac{hc}{(5\lambda /4)}-\varphi =\frac{4hc}{5\lambda }-\varphi$
Clearly, $v^{\prime }>\sqrt{\frac{5}{3}}v$