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Question: The first line of the Balmer series of \(H{{e}^{+}}\) has a wavelength of 164 nm. What is the wavele...

The first line of the Balmer series of He+H{{e}^{+}} has a wavelength of 164 nm. What is the wavelength for the series limit?
(A) 91 nm
(B) 122 nm
(C) 213 nm
(D) 107 nm

Explanation

Solution

Visible reason spectrum lies between 400 nm to 780 nm. The ultraviolet spectrum lies between 100nm to 400 nm. The balmer series is named after the Swiss teacher Johann Balmer in the year 1885, which is a series of spectral emission lines of the hydrogen atom that results from the transitions of electrons from the higher levels to the energy levels with the principal quantum number 2.

Complete step by step solution:
Rydberg found several series of spectra that would fit a more general series of spectra that would fit a more general relationship which was similar to the Balmer empirical formula come to be known as a Rydberg formula given as:
1λ=Rh[1nf2−1ni2]\dfrac{1}{\lambda }={{R}_{h}}[\dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}}]
Given in the question:
The Balmer series of He+H{{e}^{+}} has a wavelength of 164 nm
For the first line in the Balmer series the value of n1=2 and n2=3{{n}_{1}}=2\text{ and }{{\text{n}}_{2}}=3
The first equation will be:
1164=Rh[122−132]\dfrac{1}{164}={{R}_{h}}[\dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}}]
For the series limit in the Balmer series the value of n1=2 and n2=∞{{n}_{1}}=2\text{ and }{{\text{n}}_{2}}=\infty
The second equation is:
1λ2=Rh[122−1∞]\dfrac{1}{{{\lambda }_{2}}}={{R}_{h}}[\dfrac{1}{{{2}^{2}}}-\dfrac{1}{\infty }]
After dividing the equation 1 by equation 2 we can get the value of second wavelength
λ2=91.2nm{{\lambda }_{2}}=91.2nm

Hence the correct answer is option (A) i.e. the first line of the Balmer series of He+H{{e}^{+}} has a wavelength of 164 nm. What is the wavelength for the series limit is 91 nm.

Note: Balmer gave a general formulae by the trial and error method which could describe spectra from other elements while he was studying a simple spectral pattern was the lightest atom i.e. hydrogen. The individual lines in the balmer series were given particular names as alpha, beta, gamma and delta each containing a value of 3, 4, 5 and 6 respectively.