Question 496: Atoms & Nuclei

Question

The wavelength of Lyman series of hydrogen atom appears in:

Accepted Answer

$$
\begin{aligned}
& \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{\mathrm{n}^2}ight) \mathrm{n}=2,3,4, \ldots \ldots \
& \left(\frac{1}{\lambda_{\mathrm{L}}}ight)_{\max }=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{(2)^2}ight) \quad\left(\because \frac{1}{\mathrm{R}} \simeq 912 \mathop A\limits^oight) \
& \left(\lambda_{\mathrm{L}}ight)_{\max }=\frac{4}{3} \frac{\mathrm{L}}{\mathrm{R}} \
& \left(\lambda_{\mathrm{L}}ight)_{\max }=\frac{4}{3} 	imes 912 \mathop A\limits^o=4 	imes 304 \mathop A\limits^o=1216 \mathop A\limits^o \
& \left(\frac{1}{\lambda_{\mathrm{L}}}ight)_{\min }=\mathrm{R}\left(\frac{1}{(1)^2}-\frac{1}{(\infty)^2}ight) \
& \left(\lambda_{\mathrm{L}}ight)_{\min }=\frac{1}{\mathrm{R}} \simeq 912 \mathop A\limits^o
\end{aligned}$$
Range of $$\lambda$$ is $$912 \mathop A\limits^o$$ to $$1216 \mathop A\limits^o$$ which lies in U.V. region.

Answer

visible region

Answer

far infrared region

Answer

ultraviolet region

Answer

infrared region

Atoms & Nuclei

The wavelength of Lyman series of hydrogen atom appears in:

Choose the correct answer: