Question

$A$ singly ionised helium atom in an excited state $(n = 4)$ emits a photon of energy $2.6 \,eV$. Given that the ground state energy of hydrogen atom is $-13.6\, eV,$ the energy $E_{f}$ and quantum number n of the resulting state are respectively,

• A. $E_{f} = - 13.6\, eV, n = 1$
• B. $E_{f} = - 6.0\,eV, n = 3$
• C. $E_{f} = - 6.0\, eV, n = 2$
• D. $E_{f} = - 13.6\, eV, n = 2$

Answer 1

Answer: (B)

Energy of a singly ionised helium atom in nth state is $\frac{-13.6 Z^{2}}{n^{2}} ,Z=2$
So, energy levels are

In $n = 4$ state,
Energy, $E_{4}=-\frac{13.6 \times4}{16} =-3.4\,eV$
When a photon of $2.6\, eV$ is emitted, energy of atom will be
$E_{f}=-03.4-2.6=-6\,eV$
Clearly, $n = 3.$