Question

When light containing photon of energy \(2h\nu_0\) falls on a metal of work function \(h\nu_0\), electrons of velocity \(v_1\) are ejected. When photons of energy \(5h\nu_0\) is incident, velocity of electrons ejected is \(v_2\). What is the ratio \( \frac{v_1}{v_2} \)?

Hint:

In the photoelectric effect, the kinetic energy of the ejected electrons is determined by the difference between the photon energy and the work function of the metal.

Answer 1

Step 1: Use the photoelectric equation.
The photoelectric equation is given by: \[ E_{\text{photon}} = \text{Work function} + \text{Kinetic energy of electron} \] For the first case, when the photon energy is \(2h\nu_0\), the kinetic energy of the ejected electron is: \[ K_1 = 2h\nu_0 - h\nu_0 = h\nu_0 \] Thus, the velocity \(v_1\) of the electron is related to its kinetic energy by: \[ \frac{1}{2} m v_1^2 = h\nu_0 \] So: \[ v_1^2 = \frac{2h\nu_0}{m} \]
Step 2: For the second case, when the photon energy is \(5h\nu_0\), the kinetic energy of the ejected electron is:}
\[ K_2 = 5h\nu_0 - h\nu_0 = 4h\nu_0 \] Thus, the velocity \(v_2\) is related to its kinetic energy by: \[ \frac{1}{2} m v_2^2 = 4h\nu_0 \] So: \[ v_2^2 = \frac{8h\nu_0}{m} \]
Step 3: Find the ratio \( \frac{v_1}{v_2} \).
Now, we find the ratio \( \frac{v_1}{v_2} \): \[ \frac{v_1}{v_2} = \sqrt{\frac{v_1^2}{v_2^2}} = \sqrt{\frac{\frac{2h\nu_0}{m}}{\frac{8h\nu_0}{m}}} = \sqrt{\frac{2}{8}} = \frac{1}{\sqrt{4}} = \frac{1}{2} \] Thus, the ratio of the velocities is: \[ \boxed{\frac{v_1}{v_2} = \frac{1}{2}} \]