If the work function of a metal is `'phi'` and the frequency of the incident light is `'v'`, there is no emission of photoelectron if

Options

A. `v lt phi/h`
B. `v=phi/h`
C. `v gt phi/h`
D. `v gt = lt phi/h`

Answer

To solve the problem regarding the emission of photoelectrons from a metal when exposed to light of a certain frequency, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Work Function**:
The work function (φ) of a metal is the minimum energy required to remove an electron from the surface of that metal.

2. **Energy of Incident Light**:
The energy (E) of the incident light can be calculated using the formula:
\[ E = h \cdot v \]
where:
- \(E\) is the energy of the incident light,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(v\) is the frequency of the incident light.

3. **Condition for Photoelectron Emission**:
For photoelectrons to be emitted from the metal, the energy of the incident light must be greater than or equal to the work function of the metal:
\[ E \geq \phi \]
This means:
\[ h \cdot v \geq \phi \]

4. **Condition for No Emission**:
If there is no emission of photoelectrons, it implies that the energy of the incident light is less than the work function:
\[ h \cdot v < \phi \]

5. **Rearranging the Inequality**:
To express this condition in terms of frequency, we rearrange the inequality:
\[ v < \frac{\phi}{h} \]

6. **Conclusion**:
Therefore, the condition for no emission of photoelectrons is that the frequency of the incident light must be less than the ratio of the work function to Planck's constant:
\[ v < \frac{\phi}{h} \]

### Final Answer:
There is no emission of photoelectrons if:
\[ v < \frac{\phi}{h} \]