Question

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

• A. $E_f=-13.6eV,n=1$
• B. $E_f=-6.0eV,n=3$
• C. $E_f=-6.0eV,n=2$
• D. $E_f=-13.6eV,n=2$

Answer 1

Answer: (B)

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

In $n=4$ state,
Energy, $E_4=-\frac{13.6\times 4}{16}=-3.4eV$
When a photon of $2.6eV$ is emitted, energy of atom will be
$E_f=-03.4-2.6=-6eV$
Clearly, $n=3.$