Light of wavelength `lambda` falls on metal having work functions `hc//lambda_(0)` . Photoelectric effect will take place only if :

Options

A. `lambda ge(lambda_0)`
B. `lambdage 2 lambda_0`
C. `lambda le lambda_0`
D. `lambdalt lambda_0/2`

Answer

To solve the question, we need to determine the condition under which the photoelectric effect occurs when light of wavelength `lambda` falls on a metal with a work function given by `hc/lambda_0`.

### Step-by-Step Solution:

1. **Understand the Work Function**: The work function (φ₀) is the minimum energy required to eject an electron from the metal surface. In this case, it is given as:
\[ \phi_0 = \frac{hc}{\lambda_0} \]
where `h` is Planck's constant, `c` is the speed of light, and `λ₀` is a specific wavelength related to the work function.

2. **Photon Energy Calculation**: The energy (E) of a photon with wavelength `lambda` is given by the formula:
\[ E = \frac{hc}{\lambda} \]

3. **Condition for Photoelectric Emission**: For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function:
\[ E \geq \phi_0 \]
Substituting the expressions for E and φ₀, we get:
\[ \frac{hc}{\lambda} \geq \frac{hc}{\lambda_0} \]

4. **Simplifying the Inequality**: We can cancel `hc` from both sides (assuming `hc` is not zero):
\[ \frac{1}{\lambda} \geq \frac{1}{\lambda_0} \]

5. **Reciprocal of the Inequality**: Taking the reciprocal of both sides reverses the inequality:
\[ \lambda \leq \lambda_0 \]

6. **Conclusion**: The condition for the photoelectric effect to take place is that the wavelength `lambda` must be less than or equal to `lambda_0`.

### Final Answer:
The photoelectric effect will take place only if:
\[ \lambda \leq \lambda_0 \]