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  • Assertion (A): Balmer series lies in the visible region of electromagnetic spectrum. Reason   (R): <img alt="\mathrm{\frac{1}{\lambda}=R\left(\fra

Assertion (A): Balmer series lies in the visible region of electromagnetic spectrum.
Reason   (R): \mathrm{\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), where, n=3,4,5, \ldots}
 

Option: 1

If both assertion and reason are true and reason is the correct explanation of assertion
 


Option: 2

If both assertion and reason are true but the reason is not the correct explanation of assertion
 


Option: 3

If assertion is true but reason is false


Option: 4

If both assertion and reason are false


Answers (1)

best_answer

The wavelength in Baimer series is given by

\mathrm{\frac{1}{\lambda} =R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), n=3,4,5 }

\mathrm{\frac{1}{\lambda_{\max }} =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) }

\mathrm{\lambda_{\max } =\frac{36}{5 R}=\frac{36}{5 \times 1.097 \times 10^7}=6563 \AA }

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

\mathrm{\lambda_{\min } =\frac{4}{R}=\frac{4}{1.097 \times 10^7}=3646 \AA }

The wavelength \mathrm{6563 \AA\: and \: 3646 \AA} lie in visible region. Therefore, Balmer series in visible region.

Hence option 1 is correct.

Posted by

Devendra Khairwa

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