Atoms - Result Question 43
45. The ionization energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is
[1992]
(a) $3.40 eV$
(b) $1.51 eV$
(c) $0.85 eV$
(d) $0.66 eV$
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Answer:
Correct Answer: 45. (d)
Solution:
- (d) $E=E_4-E_3$
$ =-\frac{13.6}{4^{2}}-(-\frac{13.6}{3^{2}})=-0.85+1.51 $
$ =0.66 eV $
