Question:

The longest wavelength line in the Lyman series of the hydrogen spectrum is

Updated On: Jan 26, 2024
  • $ 949.8\,?$
  • $ 972.6\,?$
  • $ 1215.8\,?$
  • $ 1025.8\,?$
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The Correct Option is C

Solution and Explanation

$\frac{1}{\lambda} R_{H}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
For longest wavelength in Lyman series,
$n_{1}=1, n_{2}=2 $
$\frac{1}{\lambda}=1.09678 \times 10^{7}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right] $
$\frac{1}{\lambda}=\frac{1.09678 \times 10^{7} \times 3}{4} $
$\therefore \lambda=1.2156 \times 10^{-7} \,m $
$=1215.6\,?$
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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.