The work function of a metal is `hv_0`. Light of frequency `v` falls on this metal. The photoelectric effect will take place only if

Options

A. `vge(v_0)`
B. `vgt(2v_0)`
C. `vlt(v_0)`
D. `vlt(v_0)/2.`

Answer

To determine the condition under which the photoelectric effect will take place when light of frequency \( v \) falls on a metal with a work function of \( h\nu_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Work Function**: The work function \( \Phi_0 \) of a metal is defined as the minimum energy required to eject an electron from the surface of the metal. It is given by:
\[ \Phi_0 = h\nu_0 \]
where \( h \) is Planck's constant and \( \nu_0 \) is the threshold frequency.

2. **Energy of Incident Photon**: The energy \( E \) of a photon with frequency \( \nu \) is given by:
\[ E = h\nu \]

3. **Condition for Photoelectric Effect**: For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function of the metal. Thus, we have the condition:
\[ E \geq \Phi_0 \]

4. **Substituting the Expressions**: Substituting the expressions for energy and work function into the inequality gives:
\[ h\nu \geq h\nu_0 \]

5. **Cancelling Planck's Constant**: Since \( h \) is a positive constant, we can divide both sides of the inequality by \( h \) without changing the direction of the inequality:
\[ \nu \geq \nu_0 \]

6. **Conclusion**: Therefore, the condition for the photoelectric effect to take place is:
\[ \nu \geq \nu_0 \]

### Final Answer:
The photoelectric effect will take place only if the frequency \( \nu \) of the incident light is greater than or equal to the threshold frequency \( \nu_0 \).