NEET Physics Nuclei Questions Solved

Consider a hydrogen like atom whose energy in nth exicited state is given by En=-13.6 Z2n2 when this excited atom makes a transition from excited state to ground state, most energetic photons have energy Emax = 52.224 eV and least energetic photons have energy Emin = 1.224 eV. The atomic number of atom is
(a) 2                 (b) 5
(c) 4                 (d) None of these

(a) Maximum energy is liberated for transition En1 and minimum energy for  EnEn-1
Hence E1n2-E1=52.224 eV         ……(i)
and E1n2-E1n-12=1.224 eV…..(ii)
Solving equations (i) and (ii) we get
and E1=-54.4 eV and n = 5
Now E1=-13.6 Z212=-54.4 eV. Hence Z = 2 

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