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)

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