Target Exam

CUET

Subject

Physics

Chapter

Dual Nature of Radiation and Matter

Question:

In a photoemissive cell with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to $\frac{λ}{4}$, the speed of the fastest emitted electron will be:

Options:

2v

$\frac{v}{2}$

< 2v

> 2v

Correct Answer:

> 2v

Explanation:

As, $\frac{1}{2}mv_{max}^2=\frac{hc}{λ}-\phi_0$

$m_{max}=\sqrt{\frac{2}{m}(\frac{hc}{λ}-\phi_0)}=v$

$v'_{max}=\sqrt{\frac{2}{m}(\frac{hc}{λ/4}-\phi_0)}$

$=2\sqrt{\frac{2}{m}(\frac{hc}{λ}-\phi_0)+\frac{6}{m}\phi_0}$

> 2v