Question

When photons of energy \(h\nu\) fall on a metal plate of work function \(W_{0}\), photoelectrons of maximum kinetic energy K are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of ejected photoelectrons will be

• A. \(K+W_{0}\)
• B. \(K+h\nu\)
• C. \(K\)
• D. \(2K\)

Hint:

Write Einstein's equation for both frequencies and express the new kinetic energy using the original K and h nu.

Answer 1

The Correct Option is B

Step 1: Subtract the Two Equations
$K' - K = 2h\nu - h\nu = h\nu$, because the work function is the same in both cases.

Step 2: Solve
$K' = K + h\nu$

Step 3: Eliminate the Others
Doubling the frequency does not double K, because the work function is fixed, so $2K$ is wrong.
The result is not $K + W_0$ either, because the extra energy added is $h\nu$.

Final Answer:
The extra photon energy h nu is added to the original kinetic energy K. \[ \boxed{\text{(B) }K+h\nu} \]