Question

When the electromagnetic radiations of frequencies $4\times {10}^{15}Hz$ and $6\times {10}^{15}Hz$ fall on the same metal, in different experiments, ratio of maximum kinetic energy of electrons liberated is $1:3$. The threshold frequency for the metal is:

• A. $2\times {10}^{15}Hz$
• B. $1\times {10}^{15}Hz$
• C. $3\times {10}^{15}Hz$
• D. $1.67\times {10}^{15}Hz$

Answer 1

Answer: (C)

$3\times {10}^{15}Hz$

Let the threshold frequency is = $V_0$

$V_1=4\times {10}^{15}Hz$

$V_2=6\times {10}^{15}Hz$

now $(KE{)}_1=h{V}_1-h{V}_0$

$(KE{)}_2=h{V}_2-h{V}_0$

Given $\frac{h{V}_1-h{V}_0}{h{V}_2-h{V}_0}=\frac{1}{3}$

$3V_1-3V_0=V_2-V_0$

$\frac{(3V_1-V_2)}{2}=V_0$

$V_0=3\times {10}^{15}Hz$