The series corresponding to minimum wave- length transition in H-atom

Options

A. 3 and 1
B. 5 and 1
C. 5 and 2
D. 6 and 2

Answer

To find the series corresponding to the minimum wavelength transition in a hydrogen atom, we can use the Rydberg formula:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 1: Identify the transition for minimum wavelength
The minimum wavelength corresponds to the maximum energy transition. In hydrogen, the series that corresponds to the highest energy transitions is the Lyman series, where \( n_1 = 1 \) and \( n_2 \) can take values starting from 2 and going to infinity.

### Step 2: Calculate the wavelength for the Lyman series
For the transition from \( n_2 = 2 \) to \( n_1 = 1 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus,

\[ \lambda = \frac{4}{3R} \]

### Step 3: Calculate for higher transitions
Next, we calculate for the transition from \( n_2 = 3 \) to \( n_1 = 1 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]

Thus,

\[ \lambda = \frac{9}{8R} \]

### Step 4: Continue calculating for other transitions
For \( n_2 = 4 \) to \( n_1 = 1 \):

\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{15}{16} \right) \]

Thus,

\[ \lambda = \frac{16}{15R} \]

### Step 5: Compare wavelengths
We can see that as \( n_2 \) increases, the value of \( \lambda \) increases. Therefore, the minimum wavelength occurs at the transition from \( n_2 = 2 \) to \( n_1 = 1 \).

### Conclusion
The series corresponding to the minimum wavelength transition in the hydrogen atom is the **Lyman series**.

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