Question

Photons of energy \(1 \mathrm{eV}\) and \(2.5 \mathrm{ev}\) successively illuminate a metal, whose work function is \(0.5 \mathrm{eV}\), the ratio of maximum speed of emitted election is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(3: 1\) (D) \(1: 3\)

Short answer

The ratio of maximum speeds of emitted electrons is \(1:2\). Therefore, the correct option is (A) \(1:2\).

Step-by-step solution

Write down the photoelectric effect formula

The photoelectric effect formula is given by: \(K.E = h\nu - W\) where \(K.E\) is the kinetic energy of the emitted electrons, \(h\) is the Planck's constant, \(\nu\) is the frequency of the incident photons, and \(W\) is the work function of the metal. We can rewrite the formula in terms of energy (in eV) as: \(K.E = E_\text{photon} - W\)

Calculate the kinetic energy for each case

We are given that the energy of the incident photons for the two cases are \(1 \mathrm{eV}\) and \(2.5 \mathrm{eV}\), and the work function of the metal is \(0.5 \mathrm{eV}\). Case 1: Photon energy \(1 \mathrm{eV}\) \(K.E_1 = E_{\text{photon 1}} - W = 1\mathrm{eV} - 0.5\mathrm{eV} = 0.5\mathrm{eV}\) Case 2: Photon energy \(2.5 \mathrm{eV}\) \(K.E_2 = E_{\text{photon 2}} - W = 2.5\mathrm{eV} - 0.5\mathrm{eV} = 2\mathrm{eV}\)

Use the kinetic energy formula to calculate maximum speed

We will now use the kinetic energy formula, which states that \(K.E = \frac{1}{2}mv^2\) where \(m\) is the mass of the electron and \(v\) is its speed. For the two cases, we get: Case 1: \(K.E_1 = \frac{1}{2}m(v_1)^2 \Rightarrow v_1 = \sqrt{\frac{2K.E_1}{m}}\) Case 2: \(K.E_2 = \frac{1}{2}m(v_2)^2 \Rightarrow v_2 = \sqrt{\frac{2K.E_2}{m}}\)

Calculate the ratio of maximum speeds

To calculate the required ratio, we divide the expressions for the maximum speeds in the two cases: \(\frac{v_1}{v_2} = \frac{\sqrt{\frac{2K.E_1}{m}}}{\sqrt{\frac{2K.E_2}{m}}} = \frac{\sqrt{K.E_1}}{\sqrt{K.E_2}}\) Using the values of \(K.E_1\) and \(K.E_2\) calculated in Step 2: \(\frac{v_1}{v_2} = \frac{\sqrt{0.5\mathrm{eV}}}{\sqrt{2\mathrm{eV}}} = \frac{\sqrt{1/2}}{\sqrt{2}} = \frac{1}{\sqrt{2^2}} = \frac{1}{2}\) This ratio can be written as \(1:2\).

Answer

The ratio of maximum speeds of emitted electrons is \(1:2\). Therefore, the correct option is (A) \(1:2\).