Q 12-11-094JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
The difference between threshold wavelengths for two metal surfaces $A$ and $B$ having work function $\phi_A=9\ \text{eV}$ and $\phi_B=4.5\ \text{eV}$ in nm is (Given $hc=1242\ \text{eV nm}$)
Answer: (D) 138
$\lambda_A=\dfrac{1242}{9}=138\ \text{nm}$, $\lambda_B=\dfrac{1242}{4.5}=276\ \text{nm}$; difference $=138\ \text{nm}$.
Solution by Sreeraj P, M.Sc Physics