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 denoted as λ.

### Step-by-Step Solution:

1. **Understanding the Lyman Series**:
- The Lyman series corresponds to transitions where electrons fall to the n=1 energy level in hydrogen.
- The shortest wavelength in the Lyman series occurs when the electron transitions from n=∞ to n=1.
- The formula for the wavelength (λ) in the Lyman series is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For hydrogen (Z=1), n1=1, and n2=∞, we have:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{1^2} - 0 \right) = R \]
- Thus, the shortest wavelength in the Lyman series is:
\[ \lambda = \frac{1}{R} \]

2. **Understanding the Balmer Series**:
- The Balmer series corresponds to transitions where electrons fall to the n=2 energy level in hydrogen.
- The longest wavelength in the Balmer series occurs when the electron transitions from n=3 to n=2.
- Using the same formula for the wavelength:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For the Balmer series, n1=2 and n2=3, we have:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- Simplifying this gives:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36} \]
- Therefore, the longest wavelength in the Balmer series is:
\[ \lambda = \frac{36}{5R} \]

3. **Finding the Longest Wavelength in He⁺**:
- The He⁺ ion has Z=2.
- For the Balmer series in He⁺, the formula becomes:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- Substituting Z=2:
\[ \frac{1}{\lambda} = R \cdot 2^2 \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{20R}{36} = \frac{5R}{9} \]
- Thus, the longest wavelength in the Balmer series for He⁺ is:
\[ \lambda = \frac{9}{5R} \]

4. **Relating to the Given λ**:
- Since we know that the shortest wavelength for the Lyman series in hydrogen is λ = 1/R, we can express R in terms of λ:
\[ R = \frac{1}{\lambda} \]
- Substituting this into the expression for the longest wavelength in He⁺:
\[ \lambda = \frac{9}{5 \cdot \frac{1}{\lambda}} = \frac{9\lambda}{5} \]

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