Thereshold wavelength for sodium is ` 6 xx 10^(-7) m ` .
Then photoemission occurs for light of wavelength ` lambda ` if

Answer

To determine when photoemission occurs for light of wavelength \( \lambda \) given that the threshold wavelength for sodium is \( 6 \times 10^{-7} \, \text{m} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Threshold Wavelength**:
The threshold wavelength (\( \lambda_0 \)) is the maximum wavelength of light that can cause photoemission from a material. For sodium, this value is given as \( \lambda_0 = 6 \times 10^{-7} \, \text{m} \).

2. **Relate Wavelength to Energy**:
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light.

3. **Determine the Condition for Photoemission**:
Photoemission occurs when the energy of the incoming photon is greater than or equal to the work function (\( \phi \)) of the metal. The work function can be related to the threshold wavelength as follows:
\[ \phi = \frac{hc}{\lambda_0} \]

4. **Set Up the Inequality for Photoemission**:
For photoemission to occur, the energy of the photon must satisfy:
\[ E \geq \phi \]
Substituting the expression for energy:
\[ \frac{hc}{\lambda} \geq \frac{hc}{\lambda_0} \]

5. **Canceling Constants**:
Since \( h \) and \( c \) are constants and non-zero, we can simplify the inequality:
\[ \frac{1}{\lambda} \geq \frac{1}{\lambda_0} \]
This simplifies to:
\[ \lambda_0 \geq \lambda \]

6. **Conclusion**:
Therefore, photoemission occurs when the wavelength \( \lambda \) is less than or equal to the threshold wavelength \( \lambda_0 \):
\[ \lambda \leq 6 \times 10^{-7} \, \text{m} \]

### Final Answer:
Photoemission occurs for light of wavelength \( \lambda \) if \( \lambda \leq 6 \times 10^{-7} \, \text{m} \).