Question

A photon beam of energy 12.1eV is incident on a hydrogen atom. The orbit to

which electron of H-atom be excited is

(a) 2nd (b) 3rd (c) 4th (d) 5th



1​

Answer 1

Answer:

2nd

Explanation:

13.6 - 12.1 = 1.5

Hence the orbit is 3rd orbit

But they are asking about excited state

= 3 - 1 = 2nd

Answer 2

Answer is option c "4th"

Explanation:

• The energy of ground state electron in Hydrogen atom is -13.6 eV. Now, it absorbs 12.1 eV energy and hence, its energy increases to -13.6 + 12.1 = -0.85 eV.
• According to Bohr’s theory, energy of electron in nth orbital of hydrogenic atom/ion with atomic number Z = -13.6$\frac{Z^{2} }{n^{2} }$ eV.

• ∴ -0.85 eV = -13.6$\frac{Z^{2} }{n^{2} }$ eV.
• for hydrogen ;Z=1
• ∴ -0.85 = - 13.6 $\frac{1^{2} }{n^{2} }$  
• ∴$n^{2}$ = $\frac{13.6}{0.85}$
• ∴ $n^{2}$ = 16
• ∴ n = 4

• Therefore, the electron jumps to 4th orbit on absorbing 12.1 eV energy.