Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

A series of lines in the spectrum of atomic H lies at wavelengths 656.46 , 486.27 , 434.17 , 410.29 mm. what is the wavelength of next line in this series?

$(a)\;493.3 nm\qquad(b)\;397.2nm\qquad(c)\;593.4nm.\qquad(d)\;392.3nm$

1 Answer

The given series lies in the visible region and thus appears to be Balmer series
$n_1$ = 2
$\lambda = 410.29 \times10^{-7}$ cm
$n_2 = ?$
$\large\frac{1}{\lambda}$= $R_H[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
$\large\frac{1} {410.29\times10^{-7}}$= $109678[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
$n_2$ = 6
Next line will be obtained during the jump of electron from $7^{th}$ to $2^{nd}$ shell
$\large\frac{1}{\lambda}$= $R_H[\large\frac{1}{2^2}-\large\frac{1}{7^2}]$
= $109678[\large\frac{1}{4}-\large\frac{1}{49}]$
$\lambda = 397.2\times10^{-7}$ cm
$\lambda = 397.2$ nm
Hence answer is (b)
answered Jan 28, 2014 by sharmaaparna1

Related questions

Download clay6 mobile appDownload clay6 mobile app