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: Understanding the Concept
In the photoelectric effect, the maximum kinetic energy of an emitted electron is \(K = h\nu - \phi\), where \(\phi\) is the work function.

Step 2: Original and New Energy
Original: \(K_1 = h\nu - \phi\)
After doubling the frequency: \(K_2 = 2h\nu - \phi\)

Step 3: Compare
\(K_2 = 2(h\nu - \phi) + \phi = 2K_1 + \phi\)
Since \(\phi > 0\), \(K_2 > 2K_1\).

Step 4: Compare the options
The kinetic energy becomes more than double, option (C).

Final Answer:
The kinetic energy of the photoelectrons becomes more than double. \[ \boxed{\text{(C) }\text{more than double}} \]