Question

The wavelength of the radiation emitted is \(λ_0\) when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be \((20λ_0/x)\). The value of \(x\) is

• A. \(17\)
• B. \(21\)
• C. \(27\)
• D. \(29\)

Hint:

Use \(\frac1\lambda=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\); second excited state is \(n=3\) and third excited is \(n=4\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The ground state is \(n = 1\), the first excited is \(n = 2\), the second excited is \(n = 3\) and the third excited is \(n = 4\).

Step 2: First transition:
\(n = 3\to n = 2\): \(\frac{1}{\lambda_0} = R\left(\frac14 - \frac19\right) = \frac{5R}{36}\).

Step 3: Second transition:
\(n = 4\to n = 2\): \(\frac{1}{\lambda} = R\left(\frac14 - \frac{1}{16}\right) = \frac{3R}{16}\).

Step 4: Ratio:
\[ \lambda = \frac{16}{3R},\quad \lambda_0 = \frac{36}{5R},\quad \frac{\lambda}{\lambda_0} = \frac{16}{3}\times\frac{5}{36} = \frac{20}{27} \]
So \(\lambda = \frac{20\lambda_0}{27}\), and \(x = 27\).

Final Answer:
The value of \(x\) is \(27\), option (C). \[ \boxed{27} \]