Whenever a hydrogen atom emits a photon in the Balmer series,
it may emit another photon in the Balmer series
it must emit another photon in the Lyman series
the second photon, if emitted, will have a wavelength of about 122 nm
it may emit a second photon, but the wavelength of this photon cannot be predicted

Options

A. (i), (ii)
B. (ii), (iii)
C. (iii),(iv)
D. (ii), (iii), (iv)

Explanation

To solve the question regarding the emission of photons by a hydrogen atom in the Balmer and Lyman series, we will analyze the transitions involved and the corresponding wavelengths.

### Step-by-Step Solution:

**Step 1: Understanding the Balmer Series**
- The Balmer series involves electronic transitions in a hydrogen atom where the electron falls to the second energy level (n=2) from higher energy levels (n=3, n=4, n=5, ...).
- The wavelengths of the emitted photons can be calculated using the Rydberg formula for hydrogen.

**Step 2: Emission of a Photon in the Balmer Series**
- When a hydrogen atom emits a photon in the Balmer series, it can transition from n=3 to n=2, for example.
- This emission corresponds to a specific wavelength, which can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

**Step 3: Transition to the Lyman Series**
- After emitting a photon in the Balmer series, the atom may transition to a lower energy level, which could be in the Lyman series.
- The Lyman series involves transitions to the first energy level (n=1) from higher levels (n=2, n=3, ...).

**Step 4: Calculating the Wavelength of the Second Photon**
- If the atom transitions from n=2 to n=1 (Lyman series), we can calculate the wavelength of the emitted photon using the same Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
- Plugging in the values, we find that this transition corresponds to a wavelength of approximately 122 nm.

**Step 5: Conclusion on the Options**
- Based on the analysis:
1. The statement that it may emit another photon in the Balmer series is correct.
2. The statement that it must emit another photon in the Lyman series is also correct because transitions to n=1 are possible.
3. The wavelength of the second photon, if emitted, is indeed about 122 nm.
4. The statement that the wavelength of the second photon cannot be predicted is incorrect because we can calculate it.

### Final Answer:
- The correct options are:
- It may emit another photon in the Balmer series.
- The second photon, if emitted, will have a wavelength of about 122 nm.