Electron in hydrogen atom first jumps from third excited state to second excited state and then form second excited state to first excited state. The ratio of wavelength `lambda_(1): lambda_(2)` emitted in two cases is

Options

A. `27/5`
B. `20/7`
C. `7/5`
D. `27/20`

Answer

To find the ratio of wavelengths \( \lambda_1 : \lambda_2 \) emitted when an electron in a hydrogen atom transitions between energy levels, we can use the Rydberg formula for hydrogen spectral lines.

### Step-by-Step Solution:

1. **Identify the Energy Levels:**
- The electron transitions from the third excited state (n=4) to the second excited state (n=3) for \( \lambda_1 \).
- The electron transitions from the second excited state (n=3) to the first excited state (n=2) for \( \lambda_2 \).

2. **Apply the Rydberg Formula:**
The Rydberg formula 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, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

3. **Calculate \( \lambda_1 \):**
For the transition from \( n_i = 4 \) to \( n_f = 3 \):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_1} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_1 = \frac{144}{7R} \]

4. **Calculate \( \lambda_2 \):**
For the transition from \( n_i = 3 \) to \( n_f = 2 \):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_2} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_2 = \frac{36}{5R} \]

5. **Find the Ratio \( \lambda_1 : \lambda_2 \):**
Now we can find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{144}{7R}}{\frac{36}{5R}} = \frac{144 \times 5}{36 \times 7} \]
Simplifying this:
\[ = \frac{720}{252} = \frac{20}{7} \]
Therefore, the ratio of wavelengths emitted is:
\[ \lambda_1 : \lambda_2 = 20 : 7 \]

### Final Answer:
The ratio of wavelengths \( \lambda_1 : \lambda_2 \) is \( 20 : 7 \).