Question:

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number $Z$ of hydrogen like ion is

Updated On: Jul 18, 2024
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The Correct Option is D

Solution and Explanation

$\frac{1}{\lambda}= R Z^{2}\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)$
$R (1)^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)= R Z ^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$
$Z =2$
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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.