Question

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

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

Hint:

Use Einstein photoelectric equation to get the kinetic energies, then the speed ratio from the square root.

Answer 1

The Correct Option is B

Step 1: Write KE.
$\tfrac{1}{2}mv^2 = E - W$.

Step 2: Ratio of squares.
$\dfrac{v_1^2}{v_2^2} = \dfrac{2W}{4W} = \dfrac{1}{2}$.

Step 3: Take the root.
$v_1 : v_2 = 1 : \sqrt{2}$.

Final Answer:
Option (B). \[ \boxed{1 : \sqrt{2}} \]