Question

Light of frequency \(4\nu _{0}\) is incident on the metal of threshold frequency \(\nu _{0}\). The maximum kinetic energy of the emitted photoelectron is ( h = Planck's constant)

• A. \(3h\nu _{0}\)
• B. \(2h\nu _{0}\)
• C. \(\frac{3}{2}h\nu _{0}\)
• D. \(\frac{1}{2}h\nu _{0}\)

Hint:

Apply Einstein's photoelectric equation with the work function equal to h times the threshold frequency.

Answer 1

The Correct Option is A

Step 1: Energy balance
The photon energy is $4h\nu_0$. Of this, $h\nu_0$ is used to free the electron from the metal.

Step 2: Remaining energy
The rest appears as kinetic energy: $4h\nu_0-h\nu_0=3h\nu_0$.

Step 3: Compare the options
The value $3h\nu_0$ matches option (A). The other options would need a different incident frequency.

Final Answer:
The photoelectron has maximum kinetic energy of 3 h nu0, so option (A) is correct. \[ \boxed{\text{(A) }3h\nu_{0}} \]