Question

Energy of photon of light having two different frequencies are \(2 \mathrm{eV}\) and \(10 \mathrm{eV}\) respectively. If both are incident on the metal having work function \(1 \mathrm{eV}\), ratio of maximum velocities of emitted electron is ............ (A) \(1: 5\) (B) \(3: 11\) (C) \(2: 9\) (D) \(1: 3\)

Short answer

The ratio of the maximum velocities of the emitted electrons when photons with energies of 2 eV and 10 eV are incident on the metal with a work function of 1 eV is v1 : v2 = 1 : 3. The correct answer is (D) 1: 3.

Step-by-step solution

Understand the problem

We need to find the ratio of the maximum velocities of emitted electrons from a metal when photons with energies of 2 eV and 10 eV are incident on the metal. The work function of the metal is given as 1 eV.

Calculate the Kinetic energy of emitted electrons

According to the photoelectric effect, the excess energy of photons after overcoming the work function remains as the kinetic energy of the emitted electrons. Let the kinetic energy of emitted electrons when the incident photon has an energy of 2 eV be Ke1, and when the incident photon has an energy of 10 eV be Ke2. \(Ke1 = E_{photon1} - work\,function\) \(Ke2 = E_{photon2} - work\,function\)

Substitute given values and find Ke1 and Ke2

Substitute the values of given photon energies and work function in the equations above. \(Ke1 = 2 eV - 1 eV = 1 eV \) \(Ke2 = 10 eV - 1 eV = 9 eV \) Therefore, we have the kinetic energies of emitted electrons when photons with energies of 2 eV and 10 eV are incident on the metal.

Calculate the velocities of emitted electrons

To find the velocities of emitted electrons, we will use the formula for kinetic energy: \(Ke = \frac{1}{2} mv^2\) Where \(m\) is the mass of the electron, and \(v\) is the velocity of the electron. We want to find the ratio of the maximum velocities, so we'll need to divide both sides of the equation for Ke1 and Ke2 by \(\frac{1}{2} m\) (mass of the electron), and then take the square root of both sides: \(v1 = \sqrt{\frac{2 * Ke1}{m}}\) \(v2 = \sqrt{\frac{2 * Ke2}{m}}\)

Calculate the ratios of velocities

Now, we need to find the ratio of the velocities (v1 : v2). For this, divide both equations obtained in step 4: \(\frac{v1}{v2} = \frac{\sqrt{\frac{2 * Ke1}{m}}}{\sqrt{\frac{2 * Ke2}{m}}}\) Notice that the mass of the electron 'm' cancels out. Also, the term '2' in the numerator and denominator will cancel out. Thus, our equation becomes: \(\frac{v1}{v2} = \frac{\sqrt{Ke1}}{\sqrt{Ke2}}\) Now, substitute the values of Ke1 and Ke2 from step 3 into the equation: \(\frac{v1}{v2} = \frac{\sqrt{1 \,\text{eV}}}{\sqrt{9 \,\text{eV}}}\) \(\frac{v1}{v2} = \frac{1}{3}\) Hence, the ratio of the maximum velocities of the emitted electrons is: v1 : v2 = 1 : 3 The correct answer is (D) 1: 3.