Question

In hydrogen atom spectrum, when an electron jumps from second excited state to the first excited state, the wavelength of radiation emitted is ' \(\lambda\) '. If the electron jumps from the third excited state to the second orbit, the wavelength of radiation emitted will be \(\frac{20 \lambda}{x}\). The value of \(x\) is

• A. 18
• B. 27
• C. 21
• D. 36

Hint:

Use the Rydberg formula for each transition and divide the wavelengths.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(\dfrac{1}{\lambda}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\).

Step 2: First transition
Second excited state \(n=3\) to first excited \(n=2\):
\(\dfrac1\lambda=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}\).

Step 3: Second transition
Third excited \(n=4\) to \(n=2\):
\(\dfrac1{\lambda'}=R\left(\dfrac14-\dfrac1{16}\right)=\dfrac{3R}{16}\).

Step 4: Ratio
\(\dfrac{\lambda'}{\lambda}=\dfrac{5/36}{3/16}=\dfrac{20}{27}\). So \(\lambda'=\dfrac{20\lambda}{27}\) and \(x=27\).

Final Answer:
The value of x is 27. \[ \boxed{\text{(B) }27} \]