Question

Emission of electrons from a metal plate illuminated with monochromatic electromagnetic radiation will always take place provided

• A. The plate is positively charged
• B. The plate is negatively charged
• C. The radiation is sufficiently intense
• D. The work function of the plate is less than the energy of a single photon and the plate is uncharged

Hint:

Photoelectric emission depends on frequency (energy) of light, not intensity. Always check \(h\nu \geq \phi\).

Answer 1

The Correct Option is D

Step 1: Understand photoelectric effect.
Emission of electrons from a metal surface due to incident light is called the photoelectric effect.
For emission to occur, incident photon must have sufficient energy.

Step 2: Energy condition.
According to Einstein’s photoelectric equation:
\[ E = h\nu \] Electron emission occurs only if:
\[ h\nu \geq \phi \] where \(\phi\) is work function.

Step 3: Role of intensity.
Intensity affects number of electrons emitted, not whether emission occurs.
So option (C) is incorrect.

Step 4: Role of charge on plate.
Positive charge makes emission harder, negative charge makes it easier, but emission fundamentally depends on photon energy, not plate charge.
Thus, (A) and (B) are not the deciding condition.

Step 5: Check correct condition.
Only condition required is that photon energy exceeds work function.

Step 6: Final conclusion.
\[ \boxed{h\nu>\phi} \] Hence, correct answer is option (D).