The light emitted in the transition `n = 3 to n= 2` in hydrogen is called `H_(alpha)` light .Find the maximum work fonction a metel one have so that `H_(alpha)` light can emit photoelectrons from it

Explanation

To find the maximum work function of a metal that can emit photoelectrons when exposed to H-alpha light (which corresponds to the transition from n=3 to n=2 in hydrogen), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy of the Transition**:
The energy of the emitted photon during the transition from n=3 to n=2 can be calculated using the formula:
\[ E = 13.6 \, \text{eV} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) and \( n_2 = 3 \).

2. **Substitute the Values**:
Plugging in the values of \( n_1 \) and \( n_2 \):
\[ E = 13.6 \, \text{eV} \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ = 13.6 \, \text{eV} \left( \frac{1}{4} - \frac{1}{9} \right) \]

3. **Calculate the Fraction**:
To simplify \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

4. **Calculate the Energy**:
Now substituting back into the energy equation:
\[ E = 13.6 \, \text{eV} \times \frac{5}{36} \]
\[ = \frac{68}{36} \, \text{eV} \approx 1.89 \, \text{eV} \]

5. **Relate Energy to Work Function**:
According to the photoelectric effect, the energy of the photon must be equal to or greater than the work function \( W \) of the metal for photoelectrons to be emitted:
\[ E \geq W \]
In this case, since we are looking for the maximum work function:
\[ W_{\text{max}} = E \]

6. **Final Result**:
Therefore, the maximum work function \( W_{\text{max}} \) is:
\[ W_{\text{max}} \approx 1.89 \, \text{eV} \approx 1.9 \, \text{eV} \]

### Conclusion:
The maximum work function of the metal, so that H-alpha light can emit photoelectrons from it, is approximately **1.9 eV**.