The shortest wavelength of H-atom in Lyman series is x, then longest wavelength in Balmer series of `He^(+)` is

Options

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

Answer

To find the longest wavelength in the Balmer series of He\(^+\) given that the shortest wavelength of the H-atom in the Lyman series is \(x\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Lyman Series for Hydrogen:**
- The shortest wavelength in the Lyman series occurs when an electron transitions from \(n = \infty\) to \(n = 1\).
- The formula for the wavelength \(\lambda\) in the Lyman series is given by:
\[ \frac{1}{\lambda} = R_H \left(1 - \frac{1}{n_1^2}\right) \]
- For the shortest wavelength, \(n_1 = 1\) and \(n_2 = \infty\), so:
\[ \frac{1}{\lambda_{short}} = R_H \left(1 - 0\right) = R_H \]
- Thus, the shortest wavelength \(\lambda_{short} = x\) implies:
\[ R_H = \frac{1}{x} \]

2. **Understanding the Balmer Series for He\(^+\):**
- The longest wavelength in the Balmer series occurs when an electron transitions from \(n = 3\) to \(n = 2\).
- The formula for the wavelength \(\lambda\) in the Balmer series is given by:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For He\(^+\), the atomic number \(Z = 2\) and for the longest wavelength in the Balmer series, \(n_1 = 2\) and \(n_2 = 3\):
\[ \frac{1}{\lambda_{long}} = R_H \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

3. **Calculating the Expression:**
- Substitute \(R_H = \frac{1}{x}\):
\[ \frac{1}{\lambda_{long}} = \frac{1}{x} \cdot 4 \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Calculate the expression inside the parentheses:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
- Thus,
\[ \frac{1}{\lambda_{long}} = \frac{1}{x} \cdot 4 \cdot \frac{5}{36} = \frac{20}{36x} = \frac{5}{9x} \]

4. **Finding the Longest Wavelength:**
- Taking the reciprocal gives:
\[ \lambda_{long} = \frac{9x}{5} \]

### Final Answer:
The longest wavelength in the Balmer series of He\(^+\) is:
\[ \lambda_{long} = \frac{9x}{5} \]