Question

If the photoelectric work function of a metal is '\(\phi\)' and the frequency of incident light is '\(\nu\)', there is no emission of photoelectrons if

• A. \(\nu \lt \frac{\phi }{h}\)
• B. \(\nu =\frac{\phi }{h}\)
• C. \(\nu >\frac{\phi }{h}\)
• D. \(\nu \geq \frac{\phi }{h}\)

Hint:

Photoelectrons are emitted only if the photon energy is at least the work function, so none are emitted when the frequency is below the threshold.

Answer 1

The Correct Option is A

Step 1: Test each option
For $\nu=\dfrac{\phi}{h}$, the photon just has the work function energy, so emission with zero kinetic energy is possible. For $\nu>\dfrac{\phi}{h}$, emission occurs with positive kinetic energy.

Step 2: Why the other options fail
Option (D) $\nu\geq\dfrac{\phi}{h}$ combines the two emission cases, so it is also wrong.

Step 3: Answer
Only $\nu<\dfrac{\phi}{h}$ gives no emission, option (A).

Final Answer:
Option (A) is correct since no emission occurs below the threshold frequency. \[ \boxed{\text{(A) }\nu\lt\dfrac{\phi}{h}} \]