The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is

  • A
    $256$
  • B
    $540$
  • C
    $486$
  • D
    $626$

Explore More

Similar Questions

What is the ratio of the wavelengths of radiations emitted when an electron in a hydrogen atom jumps from the fourth orbit to the second orbit and from the third orbit to the second orbit?

What is an absorption spectrum?

Which series of the hydrogen spectrum lies in the visible region?

Ionisation potential of hydrogen atom is $13.6 eV$. When hydrogen atoms in ground state are excited by a supply of $12.1 eV$,then the number of spectral lines emitted by hydrogen atoms according to Bohr's theory is

Taking the wavelength of the first Balmer line in the hydrogen spectrum ($n = 3$ to $n = 2$) as $660\,nm$,the wavelength of the $2^{nd}$ Balmer line ($n = 4$ to $n = 2$) will be....$nm$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo