Question

The threshold frequency for a given metal is 3.6 × 10¹⁴ Hz. If monochromatic radiations of frequency 6.8 × 10¹⁴ Hz are incident on this metal, find the cut-off potential for the photoelectrons.

Answer 1

1.326 V

Answer 2

The maximum kinetic energy ($K_{max}$) of photoelectrons is given by Einstein's photoelectric equation: $K_{max} = h \nu - \phi$, where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\phi$ is the work function. The work function is related to the threshold frequency ($\nu_0$) by $\phi = h \nu_0$. Thus, $K_{max} = h \nu - h \nu_0 = h (\nu - \nu_0)$.

The cut-off potential ($V_0$) is related to $K_{max}$ by $e V_0 = K_{max}$. Equating the two expressions for $K_{max}$: $e V_0 = h (\nu - \nu_0)$

Solving for the cut-off potential: $V_0 = \frac{h (\nu - \nu_0)}{e}$

Given values:

• Threshold frequency, $\nu_0 = 3.6 \times 10^{14}$ Hz
• Frequency of incident radiation, $\nu = 6.8 \times 10^{14}$ Hz
• Planck's constant, $h = 6.63 \times 10^{-34}$ J s
• Charge of an electron, $e = 1.6 \times 10^{-19}$ C

Calculate the difference in frequencies: $\nu - \nu_0 = (6.8 \times 10^{14} \text{ Hz}) - (3.6 \times 10^{14} \text{ Hz}) = 3.2 \times 10^{14} \text{ Hz}$.

Substitute the values into the formula for $V_0$: $V_0 = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.2 \times 10^{14} \text{ Hz})}{1.6 \times 10^{-19} \text{ C}}$ $V_0 = \frac{21.216 \times 10^{-20} \text{ J}}{1.6 \times 10^{-19} \text{ C}}$ $V_0 = 1.326 \text{ V}$