The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition of an electron from the:

  • A
    Fourth Bohr orbit to the first Bohr orbit
  • B
    Fifth Bohr orbit to the second Bohr orbit
  • C
    Sixth Bohr orbit to the second Bohr orbit
  • D
    Seventh Bohr orbit to the second Bohr orbit

Explore More

Similar Questions

The ratio of longest wavelengths of the spectral lines in the Lyman and Balmer series of hydrogen spectrum is

The ratio of maximum to minimum wavelength in the Balmer series of a hydrogen atom is

Find the ratio of the wavelength of the first line of the Lyman series for a doubly ionized lithium atom $(Li^{2+})$ to the wavelength of the first line of the Lyman series for deuterium $(D)$.

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

In the Lyman series,the series limit of wavelength is $\lambda_1$. The wavelength of the first line of the Lyman series is $\lambda_2$ and in the Balmer series,the series limit of wavelength is $\lambda_3$. Then the relation between $\lambda_1$,$\lambda_2$,and $\lambda_3$ is

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