Question

The wavelength of the radiation emitted, when a Hydrogen atom electron falls from infinity to the stationary state, would be:

Rydberg constant $=1.097\times {10}^7{\mathrm{m}}^{-1}$

• A. $91\mathrm{nm}$
• B. $406\mathrm{nm}$
• C. $192\mathrm{nm}$
• D. $9.1\times {10}^{-8}\mathrm{nm}$

Answer 1

The correct option is A

$91\mathrm{nm}$

Explanation for the correct option:

(A) $\mathbf{91}\mathbf{nm}$

The wavelength of the radiation emitted is given by

$\frac{1}{\lambda }=\mathrm{R}\left[\frac{1}{{\left({\mathrm{n}}_1\right)}^2}-\frac{1}{{\left({\mathrm{n}}_2\right)}^2}\right]$

Where

• $\mathrm{\lambda }-$Wavelength
• $\mathrm{R}-$Rydberg Constant
• ${\mathrm{n}}_1-$Initial state
• ${\mathrm{n}}_2-$Final state

Here given that

• ${\mathrm{n}}_1=1$
• ${\mathrm{n}}_2=\infty$
• $\mathrm{R}=1.097\times {10}^7{\mathrm{M}}^{-1}$

Now

• $\frac{1}{\mathrm{\lambda }}=1.097\times {10}^7\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]$
• $\frac{1}{\mathrm{\lambda }}=1.097\times {10}^7{\mathrm{m}}^{-1}$
• $\mathrm{\lambda }=\frac{1}{1.097\times {10}^7}{\mathrm{m}}^{-1}$
• $\mathrm{\lambda }=9.1\times {10}^{-8}\mathrm{m}$
• $\mathrm{\lambda }=91\mathrm{nm}$

Hence, The Wavelength is $\mathbf{\lambda }\mathbf{}\mathbf{=}\mathbf{}\mathbf{91}\mathit{n}\mathit{m}$

Explanation for the incorrect options

The wavelength of radiation is $\mathbf{91}\mathbf{nm}$. Thus, options B, C, and D are incorrect.

Therefore, Option (A) $\mathbf{91}\mathbf{nm}$is correct.