Q22 of 805 Page 1

A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them?

Given: -


The wavelength of the monochromatic radiation, λ=975 A0


Ionization energy for the hydrogen atom= 13.6 eV


Formula used: -


The energy of monochromatic radiation of wavelength 975 A0 is given by the formula,



Where, E is the energy of the incident photon,


h is the Planck’s constant,


λ is the wavelength of the incident radiation and,


c is the speed of light.


So, after substituting the given values we get,



E = 12.75 eV


Also, by using the formula for energy of electron of hydrogen atom in nth shell, we can write,



Where, nfis the final principal quantum number and ni is the initial quantum number and Z is the atomic mass (for hydrogen Z=1),





Thus, the number of lines possible in the resultant spectrum is 6



The transition from n=4 to n=3 corresponds to the emission of minimum energy thus has maximum wavelength.


Conclusion: -


The number of lines possible in the resultant spectrum is 6.


The transition from n=4 to n=3 corresponds to the emission of minimum energy thus has maximum wavelength.


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