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  • The largest wavelength in the ultraviolet region of the hydrogen spectrum is . The smallest wavelength in the infrared r

The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 \mathrm{~nm}. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is:


 

Option: 1

802 nm


Option: 2

823 nm


Option: 3

1882 nm


Option: 4

1648 nm


Answers (1)

best_answer

The smallest frequency and largest wavelength in ultraviolet region will be for transition from \mathrm{n}_2=2 \text{ to } \mathrm{n}_1=1

\begin{aligned} & \therefore \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}^2}\right) \\ & \Rightarrow \frac{1}{\lambda_{\max }}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\mathrm{R}\left[1-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4} \end{aligned}

The highest frequency and smallest wavelength for infrared region will be for transition from \mathrm{n}_2=\infty \text{ to } \mathrm{n}_1=3

 \therefore\frac{1}{\lambda_{\min }}=R\left(\frac{1}{3^2}-\frac{1}{\infty}\right)=\frac{R}{9}

Divide (i) by (ii), we get

\begin{aligned} & \frac{\lambda_{\min }}{\lambda_{\max }}=\frac{3 R}{4} \times \frac{9}{R}=\frac{27}{4} \\ & \lambda_{\min }=\frac{27}{4} \times \lambda_{\max }=\frac{27}{4} \times 122 \mathrm{~nm}=823.5 \mathrm{~nm} \end{aligned}

Posted by

Suraj Bhandari

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