The ratio of wavelenghts of photons emitted when hydrogen atom de-excites from third excieted state to second excited state then de-excities form second excited state to first excited state is :

Options

A. `20//7`
B. `27//5`
C. `7//5`
D. `9//7`

Answer

To find the ratio of the wavelengths of photons emitted when a hydrogen atom de-excites from the third excited state to the second excited state and then from the second excited state to the first excited state, we can follow these steps:

### Step 1: Identify the energy levels
The hydrogen atom has energy levels given by the principal quantum number \( n \). The third excited state corresponds to \( n = 4 \), the second excited state corresponds to \( n = 3 \), and the first excited state corresponds to \( n = 2 \).

### Step 2: Write the formula for wavelength
The wavelength of the emitted photon during a transition can be derived from the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_{\text{final}}^2} - \frac{1}{n_{\text{initial}}^2} \right) \]
where \( R \) is the Rydberg constant, \( n_{\text{final}} \) is the lower energy level, and \( n_{\text{initial}} \) is the higher energy level.

### Step 3: Calculate for the first transition (from \( n = 4 \) to \( n = 3 \))
For the transition from the third excited state (\( n = 4 \)) to the second excited state (\( n = 3 \)):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (which is 144):
\[ \frac{1}{\lambda_1} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_1 = \frac{144}{7R} \]

### Step 4: Calculate for the second transition (from \( n = 3 \) to \( n = 2 \))
For the transition from the second excited state (\( n = 3 \)) to the first excited state (\( n = 2 \)):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (which is 36):
\[ \frac{1}{\lambda_2} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_2 = \frac{36}{5R} \]

### Step 5: Find the ratio of wavelengths
Now we can find the ratio of the wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{144}{7R}}{\frac{36}{5R}} = \frac{144}{7} \cdot \frac{5}{36} \]
Simplifying this:
\[ \frac{\lambda_1}{\lambda_2} = \frac{144 \cdot 5}{7 \cdot 36} = \frac{720}{252} = \frac{20}{7} \]

### Final Result
Thus, the ratio of the wavelengths of the photons emitted is:
\[ \lambda_1 : \lambda_2 = 20 : 7 \]