In hydrogen spectrum shortest wave length for lyman series line is `lamda`, then find the longest wavelength of Balmer series line in `He^+` ion spectrum

Options

A. `lamda`
B. `9/5lamda`
C. `5/9lamda`
D. `4/9lamda`

Answer

To solve the problem, we need to find the longest wavelength of the Balmer series line in the He⁺ ion spectrum, given that the shortest wavelength for the Lyman series line in hydrogen is λ.

### Step-by-Step Solution:

1. **Understanding the Lyman Series in Hydrogen**:
- The Lyman series corresponds to transitions where the electron falls to the n=1 level. The shortest wavelength (highest energy) occurs when the electron transitions from n=∞ to n=1.
- The formula for the wavelength (λ) in the Lyman series is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For hydrogen (Z=1), the shortest wavelength occurs at n1=1 and n2=∞:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \cdot 1 \cdot 1 = R \]
- Thus, the shortest wavelength λ for the Lyman series is:
\[ \lambda = \frac{1}{R} \]

2. **Understanding the Balmer Series in He⁺**:
- The Balmer series corresponds to transitions where the electron falls to the n=2 level. The longest wavelength (lowest energy) occurs when the electron transitions from n=3 to n=2.
- For He⁺ (Z=2), we apply the same formula:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, for the longest wavelength in the Balmer series, we have n1=2 and n2=3:
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

3. **Calculating the Wavelength**:
- Substitute Z=2, n1=2, n2=3 into the equation:
\[ \frac{1}{\lambda} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Calculate the difference:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
- Thus, we have:
\[ \frac{1}{\lambda} = R \cdot 4 \cdot \frac{5}{36} = \frac{20R}{36} = \frac{5R}{9} \]
- Therefore, the wavelength λ is:
\[ \lambda = \frac{9}{5R} \]

4. **Relating to λ in Hydrogen**:
- Since we know that λ (shortest wavelength in Lyman series for hydrogen) is given as λ, we can replace R with \( \frac{1}{\lambda} \):
\[ \lambda = \frac{9}{5} \cdot \lambda \]

5. **Final Answer**:
- The longest wavelength of the Balmer series line in the He⁺ ion spectrum is:
\[ \lambda_{He^+} = \frac{9}{5} \lambda \]

### Conclusion:
The answer is (B) \( \frac{9}{5} \lambda \).