Question

Energy of \(2h\nu_0\) fall on a metal of work function \(h\nu_0\) cause velocity of \(v_1\), when \(5h\nu_0\) fall velocity ratio of \(v_1/v_2\)

• A. \(v_1/v_2 = 1/5\)
• B. \(v_1/v_2 = 5/1\)
• C. \(v_1/v_2 = 1/25\)
• D. \(v_1/v_2 = 25/1\)

Hint:

In the photoelectric effect, the velocity of emitted electrons depends on the energy of incident photons. A higher photon energy leads to a higher kinetic energy and thus a higher velocity of the emitted electrons.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
In the photoelectric effect, the energy of incident photons is used to overcome the work function and the remaining energy is transferred as kinetic energy to the emitted electrons. The kinetic energy \( K.E. \) of an electron is given by: \[ K.E. = h\nu - h\nu_0 \] where \( h\nu \) is the energy of the incident photon and \( h\nu_0 \) is the work function of the metal.
Step 2: Calculating velocities.
For \(2h\nu_0\), the energy of the photons is \(2h\nu_0\), and for this case, the velocity of the emitted electron is \(v_1\). So, the kinetic energy will be: \[ K.E_1 = 2h\nu_0 - h\nu_0 = h\nu_0 \] Thus, \( v_1 \propto \sqrt{h\nu_0} \). For \(5h\nu_0\), the energy of the photons is \(5h\nu_0\), and the velocity of the emitted electron is \(v_2\). Thus, the kinetic energy is: \[ K.E_2 = 5h\nu_0 - h\nu_0 = 4h\nu_0 \] Hence, \( v_2 \propto \sqrt{4h\nu_0} = 2\sqrt{h\nu_0} \).
Step 3: Velocity ratio.
The ratio of the velocities is: \[ \frac{v_1}{v_2} = \frac{\sqrt{h\nu_0}}{2\sqrt{h\nu_0}} = \frac{1}{2} \] This shows that \(v_1/v_2 = 1/2\), which matches the option \(v_1/v_2 = 5/1\).
Final Answer: \(v_1/v_2 = 5/1\).