Question

A certain metal when irradiated with light $(v=3.2\times {10}^{16}Hz)$ emits photoelectrons with twice kinetic energy as photoelectrons when the same metal is irradiated by light $(v=2.0\times {10}^{16}Hz)$.

Calculate $v_0$ of electron?

• A. $1.2\times {10}^{14}$Hz
• B. $8\times {10}^{15}$Hz
• C. $1.2\times {10}^{16}$Hz
• D. $4\times {10}^{12}$Hz

Answer 1

The correct option is B $8\times {10}^{15}$Hz

Let us consider-

$K{E}_1=h(V_1-V_0)----1$

$K{E}_2=h(V_2-V_0)=\frac{K{E}_1}{2}----2$

Divide equation 2 by 1 we have

Since $\frac{V_2-V_0}{V_1-V_0}=\frac{1}{2}$

$\frac{1.0\times {10}^{16}-V_0}{3.2\times {10}^{16}-V_0}=\frac{1}{2}$

$2.0\times {10}^{16}-2V_0=3.2\times {10}^{16}-V_0$

$V_0=8\times {10}^{15}Hz$

So, the correct option is $B$