The threshold wavelength for a metal having work function `W_0` is `lamda_0`. What is the threshold wavelength for a metal whose work function is `W_0//2`

Options

A. `4lamda_0`
B. `2lamda_0`
C. `lamda_0//2`
D. `lamda_0//4`

Explanation

To solve the problem, we need to understand the relationship between the work function of a metal and its threshold wavelength. The threshold wavelength is the maximum wavelength of light that can cause the photoelectric effect in a metal, and it is related to the work function by the equation:

\[ W_0 = \frac{hc}{\lambda_0} \]

where:
- \( W_0 \) is the work function,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda_0 \) is the threshold wavelength.

### Step-by-Step Solution:

1. **Start with the relationship for the initial work function**:
\[ W_0 = \frac{hc}{\lambda_0} \]

2. **Now consider the new work function**:
The new work function is given as \( \frac{W_0}{2} \). Therefore, we can write:
\[ \frac{W_0}{2} = \frac{hc}{\lambda} \]
where \( \lambda \) is the new threshold wavelength we want to find.

3. **Set up the equation for the new work function**:
\[ \frac{W_0}{2} = \frac{hc}{\lambda} \]

4. **Substituting the expression for \( W_0 \)**:
From the first equation, we can substitute \( W_0 \):
\[ \frac{1}{2} \left(\frac{hc}{\lambda_0}\right) = \frac{hc}{\lambda} \]

5. **Cancel \( hc \) from both sides** (assuming \( hc \neq 0 \)):
\[ \frac{1}{2} \cdot \frac{1}{\lambda_0} = \frac{1}{\lambda} \]

6. **Rearranging gives us the new threshold wavelength**:
\[ \lambda = 2\lambda_0 \]

### Conclusion:
The threshold wavelength for a metal whose work function is \( \frac{W_0}{2} \) is \( 2\lambda_0 \).