Q) A light of frequency v is incident on a metal surface whose work function is Wo. The kinetic energy of the emitted electron is K. If the frequency of the incident light is doubled, then the Kinetic Energy of the emitted electron will be?

Q) A light of frequency v is incident on a metal surface whose work function is Wo. The kinetic energy of the emitted electron is K. If the frequency of the incident light is doubled, then the Kinetic Energy of the emitted electron will be?
a) 2k
b) more than 2k
c) between k and 2k
d) less than k

Correct Answer: b) more than 2k

Use the photoelectric equation:$$K = h\nu – W_0$$

Where:

• $K$ = kinetic energy
• $h$ = Planck’s constant
• $\nu$ = frequency
• $W_0$ = work function

Given:

$$K = h\nu – W_0$$

If frequency is doubled:$$K’ = h(2\nu) – W_0$$$$K’ = 2h\nu – W_0$$​

Now substitute $h\nu = K + W_0$:$$K’ = 2(K + W_0) – W_0$$ $$K’ = 2K + 2W_0 – W_0$$ $$K’ = 2K + W_0$$Compare with $2K$:

Since $W_0 > 0$,$$K’ = 2K + W_0 > 2K$$

Final Answer:

b) more than 2K