Estimate the difference in energy between first and second Bohr’s orbit for a hydrogen atom. At what minimum atomic number, a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with λ = 3.0 × 10 –8 m ? Which hydrogen atom like species does this atomic number correspond to?
Text Solution
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Wave number 
= 
= RH . Z2 
For hydrogen atom, Z = 1
For transition from n1 = 1 and n2 = 2
= 109677 cm–1 × 12 
= 82257.75 cm–1
E = h ν = 
Thus, E = 6.626 × 10–27 erg-sec × 3 × 1010 cm sec–1 82257.78 cm–1
= 16.3512 × 10–12 erg
= 16.3512 × 10–19 J
= 
= RH.Z2 
For the species emitting X-rays with λ = 3.0 × 10 –8 m or 3.0 × 10–6 cm
Therefore,
= 109677 cm–1 × Z2 × 
= 109677 × Z2 × 
or, Z2 = 
= 4.05 ~ 4
Z = 2
The atomic number of such species is 2. Hence it is He+ because Bohr’s theory is applicable for mono-electronic species.
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