Question

On two metal surfaces, a monochromatic light of $6 \text{ eV}$ was incident. They have ratio of their work function and maximum KE as $\frac{\Phi_1}{\Phi_2} = \frac{1}{2}$ and $\frac{(KE_{\max})_1}{(KE_{\max})_2} = \frac{2.62}{1}$. Then $\Phi_1$ and $\Phi_2$ values are respectively (in $\text{eV}$).

• A. $2.292, 4.584$
• B. $4.584, 2.292$
• C. $4.584, 9.168$
• D. $1.146, 2.292$

Hint:

When given ratios of work functions and kinetic energies, express both KE terms in relation to the work functions using $KE = E - \Phi$, and solve the resulting system of simultaneous equations.

Answer 1

The Correct Option is A

Incident energy $E = 6 \text{ eV}$.
Photoelectric equation: $KE_{\max} = E - \Phi$.
$(KE_{\max})_1 = 6 - \Phi_1$ (1).
$(KE_{\max})_2 = 6 - \Phi_2$ (2).
Given $\Phi_2 = 2\Phi_1$.
Given ratio of KE: $\frac{(KE_{\max})_1}{(KE_{\max})_2} = \frac{6 - \Phi_1}{6 - \Phi_2} = 2.62$.
Substitute $\Phi_2 = 2\Phi_1$: $\frac{6 - \Phi_1}{6 - 2\Phi_1} = 2.62$.
$6 - \Phi_1 = 2.62(6) - 2.62(2\Phi_1)$.
$6 - \Phi_1 = 15.72 - 5.24 \Phi_1$.
$5.24 \Phi_1 - \Phi_1 = 15.72 - 6$.
$4.24 \Phi_1 = 9.72$.
$\Phi_1 = \frac{9.72}{4.24} \approx 2.292 \text{ eV}$.
$\Phi_2 = 2\Phi_1 = 2 \times 2.292 = 4.584 \text{ eV}$.