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

Explanation

To solve the problem of finding the ratio of wavelengths \( \lambda_1 : \lambda_2 \) emitted when an electron in a hydrogen atom transitions between energy levels, we can follow these steps:

### Step 1: Identify the Energy Levels
- The electron transitions from the **third excited state** to the **second excited state** and then from the **second excited state** to the **first excited state**.
- In terms of quantum numbers:
- The **third excited state** corresponds to \( n = 4 \).
- The **second excited state** corresponds to \( n = 3 \).
- The **first excited state** corresponds to \( n = 2 \).

### Step 2: Calculate the Wavelength \( \lambda_1 \)
- The transition from \( n = 4 \) to \( n = 3 \) emits a photon with wavelength \( \lambda_1 \).
- The formula for the wavelength emitted during a transition is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

- For \( \lambda_1 \) (transition from \( n = 4 \) to \( n = 3 \)):
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
\[ = R_H \left( \frac{1}{9} - \frac{1}{16} \right) \]
\[ = R_H \left( \frac{16 - 9}{144} \right) = R_H \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_1 = \frac{144}{7 R_H} \]

### Step 3: Calculate the Wavelength \( \lambda_2 \)
- The transition from \( n = 3 \) to \( n = 2 \) emits a photon with wavelength \( \lambda_2 \).
- For \( \lambda_2 \):
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
\[ = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_2 = \frac{36}{5 R_H} \]

### Step 4: Calculate the Ratio \( \lambda_1 : \lambda_2 \)
- Now, we can find the ratio of the two wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{144}{7 R_H}}{\frac{36}{5 R_H}} = \frac{144 \cdot 5}{36 \cdot 7} \]
Simplifying this gives:
\[ = \frac{720}{252} = \frac{20}{7} \]
Thus, the ratio of wavelengths emitted is:
\[ \lambda_1 : \lambda_2 = 20 : 7 \]