Question

Calculate the wavelength of spectral line in lyman series corresponding to n2=3

Answer 1

Refer answer below :

Answer 2

Concept:

Hydrogen Spectrum: The hydrogen spectrum was studied systematically by Balmer. It consists of discrete bright lines on a dark background.

Lyman Series:  When an atom comes down from some higher energy level to the first energy level (lowest energy level), (that is, n₁ = 1 and n₂ = 2, 3, 4, 5, ...........), then the emitted spectral lines are obtained in the ultraviolet region.

Solution:

In Lyman Series, we have

n₁ = 1

n₂ = 3       (Given)

Now, we have find out the wavength.

We know that the relation between wavelength and energy levels is as follows -

$\frac{1}{\lambda} = R(\frac{1}{n_1^2} - \frac{1}{n_2^2} )$

where,

$\lambda = Wavelength\\R = Rydberg's \ Constant = 1.090 \times 10^7 m^{-1}$

Putting the values in the above formula, we get

$\frac{1}{\lambda} = (1.090 \times 10^7)(\frac{1}{1^2} - \frac{1}{3^2} )\\\\\frac{1}{\lambda} = (1.090 \times 10^7)(\frac{1}{1} - \frac{1}{9} )\\\\\frac{1}{\lambda} = (1.090 \times 10^7)(1 - \frac{1}{9} )\\\\\frac{1}{\lambda} = (1.090 \times 10^7)(\frac{9-1}{9} )\\\\\frac{1}{\lambda} = (1.090 \times 10^7)(\frac{8}{9} )\\\\\frac{1}{\lambda} = \frac{8.72 \times 10^7}{9} \\\\\frac{1}{\lambda} = 0.969 \times 10^7 \\\\\lambda = \frac{1}{0.969 \times 10^7} \\\lambda = 1.03 \times 10^{-7} m$

$\lambda = 103 \times 10^{-9} m\\\\\lambda = 103 \ nm$

Hence, the wavelength of spectral line in lyman series corresponding to n₂ = 3 is 103 nm.

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