The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
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n = 3 to n =2 states
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n = 3 to n = 1 states
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n = 2 to n = 1 states
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n = 4 to n =3 states
D.
n = 4 to n =3 states
Number of spectral lines obtained due to transition of electron from nth orbit to lower orbit is N = n(n-1)/2 and for maximum wavelength, the difference between the orbits of the series should be minimum
Number of spectral lines N = n (n-1)/2
= n (n-1)/2 = 6
= n2-n-12 = 0
(n-4)(n+3) = 0
n=4
Now as the first line of the series has the maximum wavelength, therefore, electrons jump from the 4th orbit to the third orbit.
