The second line of the Balmer series has a wavelength of $4861 Å$. The wavelength of the first line of the Balmer series is: (in $Å$)

  • A
    $1216$
  • B
    $6563$
  • C
    $4340$
  • D
    $4101$

Explore More

Similar Questions

$A$ $12.5\; eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Ratio of the wavelengths of the first line of the Lyman series and the first line of the Balmer series is

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

The ratio of the wavelengths of the spectral lines emitted due to transitions $3 \rightarrow 2$ and $2 \rightarrow 1$ orbits in the hydrogen atom is

$A$ hydrogen atom transitions from an excited state to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant,the principal quantum number $n$ of the excited state 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