Question

If the frequency of incident light falling on a photosensitive material is doubled, then kinetic energy of the emitted photoelectron will be:

• A. the same as its initial value
• B. two times its initial value
• C. more than two times its initial value
• D. less than two times its initial value

Hint:

Increasing the frequency of light boosts the photoelectron energy, whereas increasing intensity only increases the number of electrons emitted.

Answer 1

The Correct Option is C

Step 1: Einstein's photoelectric equation is given by: \[ KE = h f - \phi \] where \( h \) is Planck's constant, \( f \) is the frequency of the incident light, and \( \phi \) is the work function of the material. 
Step 2: When the frequency \( f \) is doubled, the new kinetic energy \( KE' \) is: \[ KE' = h (2f) - \phi = 2hf - \phi. \] Since \( KE = hf - \phi \), we can express the new kinetic energy as: \[ KE' = 2 KE + \phi. \] Thus, the new kinetic energy is more than twice the initial kinetic energy. \bigskip