Question

As per given figure A, B and C are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths (ie. \(\frac{λ_1}{λ_2}\)) is \(\frac{7}{4n}\) , then the value of n will be____.
 

Answer 1

Correct Answer: 5

We are given transitions in a hydrogen atom:

• \( \lambda_1 \) corresponds to \( n = 3 \rightarrow n = 2 \)
• \( \lambda_2 \) corresponds to \( n = 4 \rightarrow n = 3 \)
• We need to find \( n \) given that \( \frac{\lambda_1}{\lambda_2} = \frac{7}{4n} \)

Solution

1. Rydberg Formula:

The Rydberg formula is:

\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \)

2. Calculate \( \frac{1}{\lambda_1} \):

For \( \lambda_1 \), \( n_1 = 2 \) and \( n_2 = 3 \):

\( \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} \)

3. Calculate \( \frac{1}{\lambda_2} \):

For \( \lambda_2 \), \( n_1 = 3 \) and \( n_2 = 4 \):

\( \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \)

4. Calculate \( \frac{\lambda_1}{\lambda_2} \):

\( \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{\frac{1}{\lambda_1}}}{\frac{1}{\frac{1}{\lambda_2}}} = \frac{\frac{1}{\lambda_2}}{\frac{1}{\lambda_1}} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5R/36}{7R/144} = \frac{144 \times 5}{36 \times 7} = \frac{4 \times 5}{7} = \frac{20}{7} \)

5. Use the Given Ratio:

We are given \( \frac{\lambda_1}{\lambda_2} = \frac{7}{4n} \).

Therefore:

\( \frac{20}{7} = \frac{7}{4n} \)

6. Solve for \( n \):

\( 80n = 49 \)

\( n = \frac{49}{80} \)

Corrected Final Answer:

n = 49/80.