Chemistry, asked by Diya6958, 1 year ago

Calculate the wavelength of the first and last line in the balmer series of hydrogen spectrum

Answers

Answered by Divesh2409200
35
hope this answer will help you
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jdvh9548: I think. Your answer is wrong
Answered by IlaMends
51

Explanation:

The wavelength of the first line in the Balmer series of hydrogen spectrum.

n_i=3, n_f=2

\frac{1}{\lambda }=R[\frac{1}{n_i^{2}}-\frac{1}{n_f^{2}}]

\frac{1}{\lambda }=1.097\times 10^7 m^{-1}[\frac{1}{3^{2}}-\frac{1}{2^{2}}]

\lambda =-\frac{36}{5\times 1.097\times 10^7 m^{-1}}=6.56\times 10^{-7} m=-656 nm

As wavelength is cannot be negative.

The wavelength of the first line in the Balmer series of hydrogen spectrum is 656 nm.

The wavelength of the last line in the Balmer series of hydrogen spectrum.

n_i=\infty, n_f=2

\frac{1}{\lambda }=R[\frac{1}{n_i^{2}}-\frac{1}{n_f^{2}}]

\frac{1}{\lambda }=1.097\times 10^7 m^{-1}[\frac{1}{\infty ^{2}}-\frac{1}{2^{2}}]

\lambda =-\frac{4}{1.097\times 10^7 m^{-1}}=3.64\times 10^{-7} m=-364 nm

As wavelength is cannot be negative.

The wavelength of the last line in the Balmer series of hydrogen spectrum is 364 nm.

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