In a mixture of H and He^{+} gases,all H atom | vedclass

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(A) For a hydrogen-like atom,the energy is given by $E_{n} = -13.6 	imes \frac{Z^{2}}{n^{2}} \ 	ext{eV}$.For hydrogen $(Z=1)$:The first excited stat

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$4$

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(A) For a hydrogen-like atom,the energy is given by $E_{n} = -13.6 	imes \frac{Z^{2}}{n^{2}} \ 	ext{eV}$.For hydrogen $(Z=1)$:The first excited state is $n=2$. The ground state is $n=1$.Energy released during de-excitation from $n=2$ to $n=1$ is $\Delta E = E_{2} - E_{1} = -3.4 - (-13.6) = 10.2 \ 	ext{eV}$.For $He^{+}$ ion $(Z=2)$:The first excited state is $n=2$. The energy of this state is $E_{2} = -13.6 	imes \frac{2^{2}}{2^{2}} = -13.6 \ 	ext{eV}$.When the $He^{+}$ ion receives $10.2 \ 	ext{eV}$ from the hydrogen atom,its new energy becomes $E_{new} = -13.6 + 10.2 = -3.4 \ 	ext{eV}$.We equate this to the energy formula for $He^{+}$:$-3.4 = -13.6 	imes \frac{2^{2}}{n^{2}}$$n^{2} = \frac{13.6 	imes 4}{3.4} = 4 	imes 4 = 16$$n = 4$.

In a mixture of $H$ and $He^{+}$ gases,all $H$ atoms and $He^{+}$ ions are excited to their first excited states. Subsequently,on de-excitation in $H$ atoms,energy is transferred to the excited $He^{+}$ ion. The quantum number $n$ of the $He^{+}$ ion after the energy is transferred is:

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