Question

The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectrons is

• A. double the earlier value
• B. unchanged
• C. more than double
• D. less than double

Hint:

Use Einstein's equation K = h nu - phi and compare the new kinetic energy with twice the old one.

Answer 1

The Correct Option is C

Step 1: Take Numbers
Let $h\nu = 3$ units and $\phi = 1$ unit, so $K_1 = 2$ units.

Step 2: Double the Frequency
$K_2 = 6 - 1 = 5$ units, which is greater than $2K_1 = 4$ units.

Step 3: Result
The new energy is always larger than twice the old one, so option (C) is correct.

Final Answer:
Doubling the frequency more than doubles the kinetic energy. \[ \boxed{\text{(C) }\text{more than double}} \]