A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a maximum energy photon of $204\ eV$. If it makes a transition to quantum state $n$, a photon of energy $40.8\ eV$ is emitted. The value of $n$ will be

Question

Dr. Abubacker · Accepted Answer

(b) Let ground state energy (in $eV$) be $E_{1}$
Then from the given condition
$E_{20} - E_{1} = 204eV$
Or $\frac{E_{1}}{4n^{2}} - E_{1} = 204eV$
$\Rightarrow E_{1}\left( \frac{1}{4n^{2}} - 1 \right) = 204eV$ …(i)
and $E_{2n} - E_{n} = 40.8eV$
$\Rightarrow \frac{E_{1}}{4n^{2}} - \frac{E_{1}}{n^{2}} = E_{1}\left( - \frac{3}{4n^{2}} \right) = 40.8eV$ …(ii)
From equation (i) and (ii)
$\frac{1 - \frac{1}{4n^{2}}}{\frac{3}{4n^{2}}} = 5 \Rightarrow n = 2$

Q. A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a maximum energy photon of $204\ eV$. If it makes a transition to quantum state $n$, a photon of energy $40.8\ eV$ is emitted. The value of $n$ will be

Explanation

Reviewed by Dr. Abubacker

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