Arrange the following wavelenghts `(lamda)`of given emission lines of H atoms in increasing order Choose the correct option (1) `lamda_(4)ltlamda_(3)ltlamda_(2)ltlamda_(1)` (2)`lamda_(4)ltlamda_(2)ltlamda_(3)ltlamda_(1)` (3) `lamda_(1)ltlamda_(2)ltlamda_(3)ltlamda_(4)` (4) `lamda_(1)ltlamda_(3)ltlamda_(2)ltlamda_(4)`

Options

A. `n=3 overset(lamda_(1))rarr n=1`
B. `n=12 overset(lamda_(3))rarr n=10`
C. `n=5 overset(lamda_(2))rarr n=3`
D. `n=22 overset(lamda_(4))rarr n=20`

Answer

To solve the problem of arranging the wavelengths of the given emission lines of hydrogen atoms in increasing order, we will follow these steps:

### Step 1: Understand the Formula
The relationship between the wavelengths of the emission lines and the energy levels of the hydrogen atom is given by the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 2: Calculate Wavelengths
We will calculate the wavelengths for each transition:

1. **For \( \lambda_1 \) (Transition from 3 to 1)**:
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \]
\[ \lambda_1 = \frac{9}{8R_H} \]

2. **For \( \lambda_2 \) (Transition from 5 to 3)**:
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{25} \right) = R_H \left( \frac{25 - 9}{225} \right) = R_H \left( \frac{16}{225} \right) \]
\[ \lambda_2 = \frac{225}{16R_H} \]

3. **For \( \lambda_3 \) (Transition from 12 to 10)**:
\[ \frac{1}{\lambda_3} = R_H \left( \frac{1}{10^2} - \frac{1}{12^2} \right) = R_H \left( \frac{1}{100} - \frac{1}{144} \right) = R_H \left( \frac{144 - 100}{14400} \right) = R_H \left( \frac{44}{14400} \right) \]
\[ \lambda_3 = \frac{14400}{44R_H} = \frac{327.27}{R_H} \]

4. **For \( \lambda_4 \) (Transition from 22 to 20)**:
\[ \frac{1}{\lambda_4} = R_H \left( \frac{1}{20^2} - \frac{1}{22^2} \right) = R_H \left( \frac{1}{400} - \frac{1}{484} \right) = R_H \left( \frac{484 - 400}{193600} \right) = R_H \left( \frac{84}{193600} \right) \]
\[ \lambda_4 = \frac{193600}{84R_H} = \frac{2304.76}{R_H} \]

### Step 3: Compare Wavelengths
Now we will compare the calculated wavelengths:
- \( \lambda_1 = \frac{9}{8R_H} \)
- \( \lambda_2 = \frac{225}{16R_H} \)
- \( \lambda_3 = \frac{327.27}{R_H} \)
- \( \lambda_4 = \frac{2304.76}{R_H} \)

### Step 4: Arrange in Increasing Order
From the calculations, we can see that:
- \( \lambda_1 < \lambda_2 < \lambda_3 < \lambda_4 \)

Thus, the correct order of wavelengths in increasing order is:
\[ \lambda_1 < \lambda_2 < \lambda_3 < \lambda_4 \]

### Final Answer
The correct option is (3) \( \lambda_1 < \lambda_2 < \lambda_3 < \lambda_4 \).

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