Question

If the work function of a metal is 6.63 eV, then find the threshold frequency for photoelectric effect.

• A. 1.9 × 1015 Hz
• B. 1.6×1015 Hz
• C. 2 x 1016 Hz
• D. 1.2×1015 Hz

Answer 1

Answer: (B)

1.6×1015 Hz

Answer 2

The threshold frequency $\nu_0$ is given by:

$\phi_0 = h \nu_0$

Given:
- $\phi_0 = 6.63 \, \text{eV} = 6.63 \times 1.6 \times 10^{-19} \, \text{J} = 1.06 \times 10^{-18} \, \text{J}$,
- Planck’s constant $h = 6.63 \times 10^{-34} \, \text{J·s}$.

Thus, the threshold frequency is:

$\nu_0 = \frac{\phi_0}{h} = \frac{1.06 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} \, \text{Hz}.$

The correct option is (B) : 1.6×1015 Hz