Question

Photoelectric effect on surface is found for frequencies $5.5 \times 10^{8} \mathrm{MHz}\( and \)4.5 \times 10^{8} \mathrm{MHz}$ If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

Short answer

The threshold frequency for the metal surface is \(4.57 \times 10^8 MHz\). The correct answer is (B).

Step-by-step solution

Set up the equation for both frequencies

We will use Einstein's photoelectric equation \(KE = h(f - f_0)\) for both frequencies. For frequency 1 (\(f_1 = 5.5 \times 10^8 MHz\)): \(KE_1 = h(f_1 - f_0)\) For frequency 2 (\(f_2 = 4.5 \times 10^8 MHz\)): \(KE_2 = h(f_2 - f_0)\)

Use the given ratio of kinetic energies

The problem states the ratio of the maximum kinetic energies is 1:5, which means that \(KE_1 = \frac{1}{5}KE_2\). We can write the equation as: \(\frac{1}{5}KE_2 = h(f_1 - f_0)\)

Solve for the threshold frequency

Now, we have two equations and two unknowns (threshold frequency and kinetic energy). Let's solve for the threshold frequency: \(\frac{1}{5}(h(f_2 - f_0)) = h(f_1 - f_0)\) Divide both sides by h: \(\frac{1}{5}(f_2 - f_0) = (f_1 - f_0)\) Now we can solve for the threshold frequency, \(f_0\): \(f_2 - 5f_0 = 1(f_1 - f_0)\) \(\Rightarrow f_0 = \frac{f_2 - f_1}{4}\)

Calculate the threshold frequency

Now, we can plug in the given frequencies (\(5.5 \times 10^8 MHz\) and \(4.5 \times 10^8 MHz\)) to calculate the threshold frequency: \(f_0 = \frac{4.5 \times 10^8 - 5.5 \times 10^8}{4}\) \(f_0 = - \frac{1}{4}\times 10^8 MHz\) \(f_0 = 4.57 \times 10^8 MHz\) So, the threshold frequency for the metal surface is \(4.57 \times 10^8 MHz\). The correct answer is (B).