If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of `He^(+)` is:

Options

A. `(5A)/(9)`
B. `(36A)/(7)`
C. `(36A)/(5)`
D. `(9A)/(5)`

Answer

To solve the problem, we need to find the longest wavelength in the Paschen series of the helium ion (He⁺) given that the shortest wavelength in the Lyman series of the hydrogen atom is A.

### Step-by-Step Solution:

1. **Understanding the Lyman Series for Hydrogen:**
- The Lyman series corresponds to transitions where the final energy level (n1) is 1. The transitions can be from n2 = 2, 3, 4, ... to n1 = 1.
- The shortest wavelength in the Lyman series occurs when n2 approaches infinity (the transition from n = ∞ to n = 1).

2. **Using the Rydberg Formula for Hydrogen:**
- The Rydberg formula for the wavelength (λ) is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the shortest wavelength in the Lyman series:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H \]
- Given that the shortest wavelength is A, we have:
\[ R_H = \frac{1}{A} \]

3. **Understanding the Paschen Series for He⁺:**
- The Paschen series corresponds to transitions where the final energy level (n1) is 3. The transitions can be from n2 = 4, 5, 6, ... to n1 = 3.
- The longest wavelength in the Paschen series occurs when n2 is 4 (the transition from n = 4 to n = 3).

4. **Using the Rydberg Formula for He⁺:**
- For He⁺ (with atomic number Z = 2), the Rydberg formula becomes:
\[ \frac{1}{\lambda} = R_S \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R_S = Z^2 R_H = 4 R_H \) because Z = 2 for He⁺.
- For the longest wavelength in the Paschen series:
\[ \frac{1}{\lambda} = 4 R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
- Calculate the terms:
\[ \frac{1}{3^2} = \frac{1}{9}, \quad \frac{1}{4^2} = \frac{1}{16} \]
- Thus:
\[ \frac{1}{\lambda} = 4 R_H \left( \frac{1}{9} - \frac{1}{16} \right) \]
- Finding a common denominator (144):
\[ \frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144} \]
- Therefore:
\[ \frac{1}{\lambda} = 4 R_H \left( \frac{16 - 9}{144} \right) = 4 R_H \left( \frac{7}{144} \right) = \frac{28 R_H}{144} \]

5. **Substituting R_H:**
- Since \( R_H = \frac{1}{A} \):
\[ \frac{1}{\lambda} = \frac{28}{144 A} \]
- Therefore:
\[ \lambda = \frac{144 A}{28} = \frac{36 A}{7} \]

### Final Answer:
The longest wavelength in the Paschen series of He⁺ is:
\[ \lambda = \frac{36 A}{7} \]