Question

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$ If R is the Rydberg constant, then the principal quantum number n of the excited state is

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

Answer 1

Answer: (A)

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

Answer 2

According to Rydberg’s formula

$\frac{1}{\lambda} = R\left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$

Here, $n_{f} = 1,n_{i} = n$

$\therefore\frac{1}{\lambda} = R\left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right) = \frac{1}{\lambda} = R\left( 1 - \frac{1}{n^{2}} \right)............(i)$ multiplying equation (i) by $\lambda$ on both sides,

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

Or

$\frac{1}{n^{2}} = 1 - \frac{1}{\lambda R}or\frac{1}{n^{2}} = \frac{\lambda R - 1}{\lambda R}orn = \sqrt{\frac{\lambda R}{\lambda R - 1}}$