Question

Hydrogen atom from excited state comes to the ground by emitting a photon of wavelength $\lambda$ The value of principal quantum number 'n' of the excited state will be :(R : Rydberg constant)

• A. $\sqrt{\frac{\lambda R }{\lambda-1}}$
• B. $\sqrt{\frac{\lambda R}{\lambda R-1}}$
• C. $\sqrt{\frac{\lambda}{\lambda R -1}}$
• D. $\sqrt{\frac{\lambda R ^2}{\lambda R -1}}$

Answer 1

The Correct Option is B

To determine the principal quantum number \( n \) of the excited state from which a hydrogen atom transitions to the ground state, emitting a photon of wavelength \( \lambda \), we can use the Rydberg formula for hydrogen. This formula is given by:

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

Here:

• \( R \) is the Rydberg constant.
• \( \lambda \) is the wavelength of the emitted photon.
• \( n \) is the principal quantum number of the excited state.

Given that the atom transitions from an excited state to the ground state (n = 1), the formula simplifies to:

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

We need to solve for \( n \), so we rearrange the equation:

\(\frac{1}{\lambda} = R - \frac{R}{n^2}\)

Rearranging gives:

\(\frac{R}{n^2} = R - \frac{1}{\lambda}\)

Solving for \( n^2 \), we get:

\(n^2 = \frac{R}{R - \frac{1}{\lambda}} = \frac{R}{\frac{R\lambda - 1}{\lambda}} = \frac{\lambda R}{\lambda R - 1}\)

Taking the square root, we find:

\(n = \sqrt{\frac{\lambda R}{\lambda R - 1}}\)

Thus, the principal quantum number \( n \) of the excited state is \(\sqrt{\frac{\lambda R}{\lambda R - 1}}\), which matches the provided correct answer.