PYQ NEET- Dual Nature Of Matter And Radiation L-4

When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_s}{2}$ and $\frac{V_s}{2}$ respectively. The threshold frequency for this metal is#

A) $2\ \mathrm{v}$

B) $3,v$

C) $\frac{2}{3} \mathrm{v}$

D) $\frac{3}{2} \mathrm{v}$

Answer: $\frac{3}{2} \mathrm{v}$#

Solution:#

Since $k_{\max }=\frac{e V_s}{h}=v-\frac{\phi}{h}$ $$ \begin{alignedat} & \frac{e V_s}{2}=h v-h v_0 \ldots \ldots . \text { (i) } \ & e V_s=\frac{h v}{2}-h v_0 \ldots \ldots . \text { (ii) } \ & \frac{1}{2}\left[\frac{h v}{2}-h v_0\right]=\frac{1}{2}h v-\frac{1}{2}h v_0 \ & \Rightarrow h v_0-\frac{h v_0}{2}=h v-\frac{h v}{4} \ & \Rightarrow \frac{h v_0}{2}=\frac{3 h v}{4} \ & v_0=\frac{3 v}{2} \end{aligned} $$

• Language of question is wrongly framed. The values of stopping potentials should be interchanged.