Question

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

Short answer

The threshold frequency for the metal surface is \(f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\}\).

Step-by-step solution

Write down the given information

We are given the frequencies \(f_1\), \(f_2\) and the ratio of maximum kinetic energies \(1:K\). Let the maximum kinetic energy at frequency \(f_1\) be \(E_1\) and at frequency \(f_2\) be \(E_2\). Based on the given information: \[ \frac{E_1}{E_2} = \frac{1}{K} \]

Use the photoelectric effect formula

According to the photoelectric effect formula: For frequency \(f_1\), \[ E_1 = h(f_1 - f_0) \] For frequency \(f_2\), \[ E_2 = h(f_2 - f_0) \]

Plug energy ratio into the equations

We can plug the energy ratio into the equations: \[ \frac{E_1}{E_2} = \frac{h(f_1 - f_0)}{h(f_2 - f_0)} = \frac{1}{K} \]

Solve for the threshold frequency

Now we will solve the equation for the threshold frequency \(f_0\): \[ \frac{f_1 - f_0}{f_2 - f_0} = \frac{1}{K} \] Multiply both sides by \(K(f_2 - f_0)\): \[ K(f_1 - f_0) = f_2 - f_0 \] Now, solve for \(f_0\): \[ f_0 = \frac{Kf_1 - f_2}{K - 1} \] So the threshold frequency for the metal surface is: \[ f_0 = \left\\{\left(Kf_1 - f_2\right) / (K - 1)\right\\} \] The correct answer is (B).