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  • If be the lowest wavelength of the Lyman series then the lowest wavelength of the Balmer series will be

If 917 A^0 be the lowest wavelength of the Lyman series then the lowest wavelength of the Balmer series will be
_________ A^0.

Option: 1

3668


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

\begin{array}{l}{\text { Lyman series: } \frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right), n=2,3,4, \ldots} \\ \\ {\text { Balmer series: } \frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right), n=3,4,5, \ldots}\\ \\ \end{array}

 

The expression for wavelength is written as \frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]
The last line is the line of shortest wavelength or highest energy. When \mathrm{n}_2=\infty, we get the last line wavelength.

\begin{aligned} & \frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}\right] \\ & \lambda_{\text {Series limit }}=\frac{\mathrm{n}_1^2}{\mathrm{RZ}^2} \end{aligned}
For Lyman series, n_1=1 and for Balmer series n_1=2.
Therefore, \frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\frac{\mathrm{n}_{1 \text { Lyman }}^2}{\mathrm{n}_{1 \text { Balmer }}^2}=\frac{1^2}{2^2}=\frac{1}{4}
 

 

\lambda_{Balmer}=917 \times 4=3668

Posted by

Kshitij

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