Question

a spectral line of hydrogen with wavelength as 4938 angstrom belongs to which series ​

Answer 1

It belongs to balmer series

Answer 2

Thus, the series is the Balmar series.

Given:

A spectral line of hydrogen with a wavelength of 4938 angstroms.

To find:

The series.

Solution:

The wavelength of a spectral line is given by the below formula-

Here lambda is the wavelength having value $4938×10^{-10}m$.

R is a constant having value $1.097×10^{7}m$.

Putting the values we have-

$\frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\\\frac{1}{{4938 \times {{10}^{ - 10}}}} = 1.097 \times {10^7}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]\\\frac{1}{{4938 \times {{10}^{ - 3}}}} = \frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}\\\frac{1}{{5.4169}} = \frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}$

For n1=1, the series is Lyman.

${n_1} = 1\\\frac{1}{{5.4169}} = \frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}\\n_2=1.23$

Thus n2 is not an integer value so, the series is not Lyman.

Again for n1=2, the series is Balmar

${}{n_1} = 2\\\frac{1}{{5.4169}} = \frac{1}{{{2^2}}} - \frac{1}{{n_2^2}}\\{n_2} = 4$

Thus n2 is an integer value.

Thus, the series is the Balmar series.

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