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. `(7)/(20)`
B. `(20)/(7)`
C. 5
D. 20

Answer

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

### Step 1: Identify the Energy Levels
The energy levels of a hydrogen atom are given by the principal quantum number \( n \). The states mentioned in the question are:
- Third excited state corresponds to \( n = 4 \)
- Second excited state corresponds to \( n = 3 \)
- First excited state corresponds to \( n = 2 \)

### Step 2: Write the Formula for Wavelength
The wavelength \( \lambda \) of the emitted photon during a transition from a higher energy level \( n_i \) to a lower energy level \( n_f \) can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant.

### Step 3: Calculate Wavelength for the First Transition
For the transition from \( n = 4 \) to \( n = 3 \):
- Here, \( n_f = 3 \) and \( n_i = 4 \).
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda_1} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_1 = \frac{144}{7R} \]

### Step 4: Calculate Wavelength for the Second Transition
For the transition from \( n = 3 \) to \( n = 2 \):
- Here, \( n_f = 2 \) and \( n_i = 3 \).
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_2} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_2 = \frac{36}{5R} \]

### Step 5: Calculate the Ratio of Wavelengths
Now, we need to find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{144}{7R}}{\frac{36}{5R}} = \frac{144}{7R} \cdot \frac{5R}{36} \]
The \( R \) cancels out:
\[ = \frac{144 \cdot 5}{7 \cdot 36} \]
Calculating this:
\[ = \frac{720}{252} = \frac{20}{7} \]

### Final Answer
The ratio of the wavelengths of the photons emitted during the two transitions is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{20}{7} \]