The shortest wavelength of Balmer series of H-atom is

Options

A. `4/R`
B. `36/(5R)`
C. `1/R`
D. `3/(4R)`

Answer

To find the shortest wavelength of the Balmer series of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions where the final energy level (nf) is 2. The initial energy level (ni) can be any level greater than 2, but for the shortest wavelength, we take ni to be infinity (the highest energy level).

### Step 2: Use the Wavelength Formula
The formula to calculate the wavelength (λ) of the emitted light during a transition in a hydrogen atom is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step 3: Substitute the Values
For the shortest wavelength in the Balmer series:
- \( n_f = 2 \)
- \( n_i = \infty \)

Substituting these values into the formula:

\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

### Step 4: Solve for Wavelength
Now, we can solve for λ:

\[ \lambda = \frac{4}{R} \]

### Step 5: Conclusion
Thus, the shortest wavelength of the Balmer series of the hydrogen atom is:

\[ \lambda = \frac{4}{R} \]

### Final Answer
The shortest wavelength of the Balmer series of H-atom is \( \frac{4}{R} \).

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