The ${\rm{\lambda }}$ for ${H_\alpha }$ line of Balmer series is 6500 Å. Thus ${\rm{\lambda }}$ for ${H_\beta }$ line of Balmer series is :

JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Medium

1 of 1

Application +4 | -1

The

\mathrm{\lambda }

for

{H_\alpha }

line of Balmer series is  6500 Å. Thus

\mathrm{\lambda }

for

{H_\beta }

line of Balmer series is :

View answer and explanation

Correct option: A

4814 Å

Option A (4814 Å) is correct because it is derived from the Rydberg formula for the Balmer series, relating wavelengths to energy level transitions.

Practice All Chapter MCQs →

n^{\text {th }}

\mathrm{A}

\times 10^{12}

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

\ell = 1,\,\,m = 0

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.