Question

Hydrogen atom from $n^{th}$ energy state comes to the ground state by emitting a photon of wavelength $\lambda$. The value of ‘n’ will be: (R: Rydberg constant)

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

Answer 1

Answer: (C)

$\sqrt{\frac{\lambda R}{\lambda R - 1}}$

Answer 2

The problem asks us to find the principal quantum number 'n' of the initial energy state from which a hydrogen atom transitions to the ground state, emitting a photon of wavelength $\lambda$.

We use the Rydberg formula for the wavelength of light emitted during an electronic transition in a hydrogen atom: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$ where:

• $\lambda$ is the wavelength of the emitted photon.
• R is the Rydberg constant.
• $n_i$ is the principal quantum number of the initial energy state.
• $n_f$ is the principal quantum number of the final energy state.

Given:

• The atom comes from the $n^{th}$ energy state, so $n_i = n$.
• The atom comes to the ground state, so $n_f = 1$.

Substitute these values into the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$ $\frac{1}{\lambda} = R \left( 1 - \frac{1}{n^2} \right)$

Now, we need to solve for 'n'. Multiply both sides by $\lambda$: $1 = \lambda R \left( 1 - \frac{1}{n^2} \right)$

Divide both sides by $\lambda R$: $\frac{1}{\lambda R} = 1 - \frac{1}{n^2}$

Rearrange the equation to isolate $\frac{1}{n^2}$: $\frac{1}{n^2} = 1 - \frac{1}{\lambda R}$

Combine the terms on the right-hand side: $\frac{1}{n^2} = \frac{\lambda R - 1}{\lambda R}$

Take the reciprocal of both sides to find $n^2$: $n^2 = \frac{\lambda R}{\lambda R - 1}$

Finally, take the square root of both sides to find 'n': $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$

Comparing this result with the given options, we find that it matches option (3).