The ratio of minimum wavelengths of Lyman and Balmer series will be
- 1.25
- 0.25
- 5
- 10
The Correct Option is B
Solution and Explanation
$\lambda \propto n^{2}$
$\therefore \frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\left(\frac{1}{2}\right)^{2}$
$\frac{\lambda_{L}}{\lambda_{B}}=\frac{1}{4}=0.25$
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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.