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A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

Grade:12th pass

1 Answers

Samyak Jain
333 Points
5 years ago
We know that number of spectral lines can be calculated by \Deltan(\Deltan + 1)/2 which is 6 here.
So, \Deltan(\Deltan + 1)/2  =  6  or  \Deltan(\Deltan + 1) = 12  i.e.  \Deltan = 3.
This shows that electrons from ground state of hydrogen atoms must have been excited to 4th orbit.
Energy of nth orbit of hydrogen atom is (– 13.6) / n2 eV.
\therefore absolute difference in energy of 1st and 4th orbits is (13.6)(1/12 – 1/42) = 12.75 eV.
Energy of incident radiation is E = 12.75 eV. We can calculate its wavelength by E = hc/\lambda,
h = Planck‘s constant, c = speed of light, \lambda = walength of photon.
Or directly we know that E = 12400/\lambda, where E is in eV and \lambda is in angstrom.
So, 12.75 = 12400/\lambda  i.e. \lambda = 12400/12.75  \approx  972.55 angstrom

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