Question

Energy of the incident photon on the metal surface is ' 3W ' and then ' 5W ', where 'W' is the work function for that metal. The ratio of velocities of emitted photoelectrons is

• A. \( 1 : \sqrt{2} \)
• B. \( 1 : 1 \)
• C. \( 1 : 2 \)
• D. \( 1 : 4 \)

Hint:

In problems related to the photoelectric effect, remember that the velocity of emitted electrons is related to the square root of the energy.

Answer 1

The Correct Option is A

Step 1: Understanding the photoelectric effect.
The energy of the photon is given as \( E = h\nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency. The velocity \( v \) of the emitted photoelectrons is related to the energy by the equation \( E = \frac{1}{2}mv^2 \).

Step 2: Analyzing the energy difference.
Given that the energy of the photon changes from \( 3W \) to \( 5W \), the energy difference is \( 5W - 3W = 2W \). Since energy is proportional to the square of velocity, the ratio of velocities of emitted photoelectrons will be \( \frac{v_2}{v_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{\frac{5W}{3W}} = \sqrt{ \frac{5}{3}} = \sqrt{2} \).

Step 3: Conclusion.
The correct answer is (A) \( 1 : \sqrt{2} \), as the ratio of velocities is based on the square root of the energy ratio.