In terms of Rydberg constant `R`, the shortest wavelength in the Balmer series of the hydrogen , atom spestrum will have wavelength

Options

A. `1//R`
B. `4//R`
C. `3//2 R`
D. `9//R`

Explanation

To find the shortest wavelength in the Balmer series of the hydrogen atom spectrum in terms of the Rydberg constant \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rydberg Formula**:
The Rydberg formula for the wavelength of light emitted during electronic transitions in hydrogen is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

2. **Identify the Balmer Series**:
In the Balmer series, the transitions end at the \( n_f = 2 \) level. The initial level \( n_i \) can be any level greater than 2.

3. **Determine the Shortest Wavelength**:
The shortest wavelength corresponds to the transition from the highest energy level (which approaches infinity, \( n_i \to \infty \)) to \( n_f = 2 \). Thus, we set:
- \( n_f = 2 \)
- \( n_i = \infty \)

4. **Substitute into the Rydberg Formula**:
Plugging these values into the Rydberg formula:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
Simplifying this gives:
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

5. **Solve for Wavelength \( \lambda \)**:
Taking the reciprocal to find \( \lambda \):
\[ \lambda = \frac{4}{R} \]

### Conclusion:
The shortest wavelength in the Balmer series of the hydrogen atom spectrum is:
\[ \lambda = \frac{4}{R} \]