Question

A singly ionized 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 $\left(E_f\right)$ and quantum number $\left(n\right)$ 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)

$E_f=-6.0eV$, $n=3$

From the Bohr's theory of single electron species, we have the Energy of the state (quantum number = n) and atomic number = Z as

$E=-13.6\frac{Z^2}{n^2}$

Thus, Energy of $4^{th}$ state in Helium ion is $E_4^{H{e}^{+}}=-13.6\times \frac{2^2}{4^2}=-3.4eV$

Energy of the emitted photon is $E_p=2.6eV$

Energy after emitting the photon is $E_f=E_4^{H{e}^{+}}-E_p=-3.4-2.6=-6eV$

The quantum number is given by $n=(\frac{-13.6\times Z^2}{E_n}{)}^{\frac{1}{2}}=\sqrt{\frac{-13.6\times 4}{-6}}=\sqrt{9}=3$