Q 12-11-147JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
When a photon of energy $4.0$ eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ eV and de Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by a photon of energy $4.50$ eV is $T_B = (T_A - 1.5)$ eV. If the de Broglie wavelength of these photoelectrons is $\lambda_B = 2\lambda_A$, then the work function of metal B is:
Answer: (A) $4$ eV
$\lambda = \dfrac{h}{\sqrt{2mT}}$, so $\lambda_B = 2\lambda_A$ means $T_B = \dfrac{T_A}{4}$.
$$T_A - 1.5 = \frac{T_A}{4} \Rightarrow T_A = 2\ \text{eV},\ T_B = 0.5\ \text{eV}$$
$$\phi_B = 4.5 - 0.5 = 4\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics