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Question:

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is ______.

Solution:

ΔE21 = 13.6 × z2[1/4 - 1/1] = 13.6 × z2[3/4]
ΔE32 = 13.6 × z2[1/4 - 1/9] = 13.6 × z2[5/36]
ΔE21 = ΔE32 + 74.8
13.6 × z2[3/4] = 13.6 × z2[5/36] + 74.8
13.6 × z2[3/4 - 5/36] = 74.8
z2 = 9
z = +3 (z can't be -ve)