Question

Two identical photo-cathodes receive light of frequencies $v_1$ and $v_2$. If the velocity of the photoelectrons (of mass m) coming out are $v_1$ and $v_2$ respectively, then

• A. $v_1-v_2={\left[\frac{2h}{m}(v_1-v_2)\right]}^{1/2}$
• B. $v_1+v_2={\left[\frac{2h}{m}(v_1+v_2)\right]}^{1/2}$
• C. $v_1^2+v_2^2=\frac{2h}{m}(v_1+v_2)$
• D. $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$

Answer 1

The correct option is D $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$

Using Einstein photo electric equation,

$E=w_0+k_{max}$

${hf}_1=w_e+\frac{1}{2}{mv}_1^2⟶\left(i\right)$

${hf}_2=w_0+\frac{1}{2}{mv}_2^2⟶\left(ii\right)$

$\Rightarrow h(f_1-f_2)=\frac{1}{2}m(v_1^2-v_2^2)$

$\Rightarrow (v_1^2-v_1^2)=\frac{2h}{m}(f_1-f_2)$