A singly ionised helium atom in an excited st | vedclass

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(B) The energy of a hydrogen-like atom in the $n$th state is given by $E_n = -13.6 \frac{Z^2}{n^2} \,eV$.For a singly ionised helium atom,the atomic n

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

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(B) The energy of a hydrogen-like atom in the $n$th state is given by $E_n = -13.6 \frac{Z^2}{n^2} \,eV$.For a singly ionised helium atom,the atomic number $Z = 2$.Therefore,the energy of the state $n=4$ is:$E_4 = -13.6 	imes \frac{2^2}{4^2} = -13.6 	imes \frac{4}{16} = -3.4 \,eV$.When the atom emits a photon of energy $2.6 \,eV$,its final energy $E_f$ will be:$E_f = E_4 - 2.6 \,eV = -3.4 \,eV - 2.6 \,eV = -6.0 \,eV$.Now,we find the quantum number $n$ for this energy level using $E_n = -13.6 \frac{Z^2}{n^2}$:$-6.0 = -13.6 	imes \frac{4}{n^2} \implies n^2 = \frac{13.6 	imes 4}{6.0} \approx 9.06 \approx 9$.Thus,$n = 3$.The final energy is $-6.0 \,eV$ and the quantum number is $n=3$.

$A$ 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,

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