Question:

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :

Updated On: Sep 13, 2024
  • 1
  • 4
  • 0.5
  • 2
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The Correct Option is B

Solution and Explanation

For last Balmer series
$\frac{1}{\lambda_{b}}=R\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right]$
$\lambda_{b}=\frac{4}{R}$
For last Lyman series
$\frac{1}{\lambda_{l}}=R\left[\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right]$
$\lambda_{I}=\frac{1}{R}$
$\frac{\lambda_{b}}{\lambda_{l}}=\frac{\frac{4}{R}}{\frac{1}{R}}$
$\frac{\lambda_{b}}{\lambda_{l}}=4$
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Concepts Used:

Atomic Spectra

The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an electron making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states.

Read More: Atomic Spectra

Spectral Series of Hydrogen Atom

Rydberg Formula:

The Rydberg formula is the mathematical formula to compute the wavelength of light.

\[\frac{1}{\lambda} = RZ^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})\]

Where,

R is the Rydberg constant (1.09737*107 m-1)

Z is the atomic number

n is the upper energy level

n’ is the lower energy level

λ is the wavelength of light

Spectral series of single-electron atoms like hydrogen have Z = 1.

Uses of Atomic Spectroscopy:

  • It is used for identifying the spectral lines of materials used in metallurgy.
  • It is used in pharmaceutical industries to find the traces of materials used.
  • It can be used to study multidimensional elements.