Radiation coming from the transition n = 2 to | vedclass

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(A) The energy of a photon emitted during a transition in a hydrogen-like atom is given by $E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \rig

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$n = 2 	o n = 4$

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(A) The energy of a photon emitted during a transition in a hydrogen-like atom is given by $E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.For hydrogen $(Z=1)$ transitioning from $n=2$ to $n=1$,the energy is $E = 13.6 	imes 1^2 	imes \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 	imes \frac{3}{4} 	ext{ eV}$.For $He^+$ ions $(Z=2)$,the energy required for a transition from $n_i$ to $n_f$ is $E = 13.6 	imes 2^2 	imes \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} ight) = 13.6 	imes 4 	imes \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} ight) 	ext{ eV}$.We equate the energies: $13.6 	imes \frac{3}{4} = 13.6 	imes 4 	imes \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} ight)$.This simplifies to $\frac{3}{16} = \frac{1}{n_i^2} - \frac{1}{n_f^2}$.If the ion is in the $n=2$ state $(n_i=2)$,then $\frac{1}{4} - \frac{1}{n_f^2} = \frac{3}{16}$.$\frac{1}{n_f^2} = \frac{1}{4} - \frac{3}{16} = \frac{4-3}{16} = \frac{1}{16}$.Thus,$n_f^2 = 16$,which means $n_f = 4$.Therefore,the transition is $n = 2 	o n = 4$.

Radiation coming from the transition $n = 2$ to $n = 1$ of hydrogen atoms falls on $He^+$ ions in $n = 1$ and $n = 2$ states. The possible transition of helium ions as they absorb energy from the radiation is:

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