Whenever a photon is emitted by hydrogen in balmer series it is followed by another in lyman series what wavelength does this latter photon correspond to?

Explanation

To solve the problem of determining the wavelength of the photon emitted in the Lyman series after a photon in the Balmer series is emitted by hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Transition in the Lyman Series**:
The Lyman series corresponds to transitions where an electron falls to the n=1 energy level. In this case, since the question states that the photon in the Lyman series follows a photon in the Balmer series, we know that the transition is from n=2 to n=1.

2. **Use the Rydberg Formula**:
The wavelength (\( \lambda \)) of the emitted photon can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Substitute the Values**:
For the transition from n=2 to n=1:
- \( n_1 = 1 \)
- \( n_2 = 2 \)
- The Rydberg constant \( R \) is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).

Plugging in these values into the formula:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

4. **Calculate the Terms**:
Calculate \( \frac{1}{1^2} - \frac{1}{2^2} \):
\[ \frac{1}{1} - \frac{1}{4} = 1 - 0.25 = 0.75 \]

5. **Final Calculation**:
Now substitute this back into the equation:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.75 \]
\[ \frac{1}{\lambda} = 0.82275 \times 10^7 \, \text{m}^{-1} \]

6. **Find the Wavelength**:
Now take the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{0.82275 \times 10^7} \approx 1.215 \times 10^{-7} \, \text{m} \]
Converting to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda \approx 122 \, \text{nm} \]

### Final Answer:
The wavelength of the photon emitted in the Lyman series is approximately **122 nm**.