Question

The work functions of two metals ($M_A$ and $M_B$) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of 2.642 : 1. The work functions (in eV) of $M_A$ and $M_B$ are respectively.

• A. 2.3, 4.6
• B. 3.1, 6.2
• C. 1.4, 2.8
• D. 1.5, 3.0

Hint:

Einstein's Photoelectric Equation: $h\nu = \phi + K_{max}$.

Answer 1

The Correct Option is A

Let $\phi_A = \phi$ and $\phi_B = 2\phi$.
$K_A = 6 - \phi$.
$K_B = 6 - 2\phi$.
Ratio $\frac{K_A}{K_B} = \frac{6-\phi}{6-2\phi} = 2.642$.
$6 - \phi = 2.642(6 - 2\phi) = 15.852 - 5.284\phi$.
$4.284\phi = 9.852$.
$\phi = \frac{9.852}{4.284} \approx 2.3 \text{ eV}$.
$\phi_A = 2.3 \text{ eV}$, $\phi_B = 4.6 \text{ eV}$.