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(C) The total energy of the system before collision is the kinetic energy of the free electron,as the potential energy at a large distance is $0$. Tot

Vedclass · Accepted Answer

$1.45 	imes 10^{9} \, MHz$

Vedclass · Answer

(C) The total energy of the system before collision is the kinetic energy of the free electron,as the potential energy at a large distance is $0$. Total energy $E_i = 2.6 \, eV + 0 = 2.6 \, eV$.The hydrogen atom is formed in the first excited state $(n=2)$. The energy of the hydrogen atom in the $n$-th state is given by $E_n = -13.6/n^2 \, eV$. For $n=2$,$E_f = -13.6/4 = -3.4 \, eV$.The energy released as a photon is the difference between the initial and final total energy: $\Delta E = E_i - E_f = 2.6 - (-3.4) = 6.0 \, eV$.Convert the energy to Joules: $\Delta E = 6.0 	imes 1.6 	imes 10^{-19} \, J = 9.6 	imes 10^{-19} \, J$.Using the relation $\Delta E = hf$,the frequency $f = \Delta E / h = (9.6 	imes 10^{-19}) / (6.6 	imes 10^{-34}) \approx 1.45 	imes 10^{15} \, Hz$.Since $1 \, MHz = 10^6 \, Hz$,the frequency in $MHz$ is $1.45 	imes 10^{15} / 10^6 = 1.45 	imes 10^9 \, MHz$.

$A$ free electron of $2.6 \, eV$ energy collides with a $H^+$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. $(h = 6.6 \times 10^{-34} \, J \cdot s)$

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