Question

The wavelength of the radiation emitted is $\lambda_0$ 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 $\frac{20}{x} \lambda_0$ The value of $x$ is

Answer 1

Correct Answer: 27

To solve this problem, we need to use the concept of energy levels in a hydrogen atom. The energy difference between two levels \( n_1 \) and \( n_2 \) in a hydrogen atom can be calculated using the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( \lambda \) is the wavelength of emitted radiation, and \( R_H \) is the Rydberg constant.

Step 1: Calculate \( \lambda_0 \)
For an electron jumping from the second excited state (\( n=3 \)) to the first excited state (\( n=2 \)), the wavelength is \(\lambda_0\). Thus: \[ \frac{1}{\lambda_0} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \cdot \frac{5}{36} \]

Step 2: Calculate new wavelength
When the electron jumps from the third excited state (\( n=4 \)) to the second orbit (\( n=2 \)): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \cdot \frac{3}{16} \]

Step 3: Relate \( \lambda \) and \( \lambda_0 \)
Given \(\lambda = \frac{20}{x} \lambda_0\), equate the two expressions: \[ \frac{1}{\frac{20}{x}\lambda_0} = R_H \cdot \frac{3}{16} \]
Replacing \(\lambda_0\): \[ \frac{x}{20\lambda_0} = R_H \cdot \frac{3}{16} \]
Using the expression for \( \lambda_0 \): \[ \frac{x}{20} \cdot R_H \cdot \frac{5}{36} = R_H \cdot \frac{3}{16} \]

Simplifying and solving for \( x \): \[ \frac{x}{20} \cdot \frac{5}{36} = \frac{3}{16} \rightarrow \frac{5x}{720} = \frac{3}{16} \rightarrow 5x = 135 \rightarrow x = 27 \]

Therefore, \( x = 27 \), which is within the expected range (27,27).