A photon collides with a stationary hydrogen | vedclass

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(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \ eV$. The first excited state energy is $E_2 = -3.4 \ eV$. The energy difference

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One photon of energy $10.2 \ eV$ and an electron of energy $1.4 \ eV$

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(C) $1$. The ground state energy of a hydrogen atom is $E_1 = -13.6 \ eV$. The first excited state energy is $E_2 = -3.4 \ eV$. The energy difference is $\Delta E = E_2 - E_1 = 10.2 \ eV$.$2$. When a $10.2 \ eV$ photon hits the atom,it is absorbed,exciting the electron to the $n=2$ state. After a microsecond,the electron de-excites,emitting a $10.2 \ eV$ photon.$3$. When the second photon of $15 \ eV$ hits the atom,since $15 \ eV > 13.6 \ eV$ (the ionization energy of hydrogen),the electron is ejected from the atom.$4$. The kinetic energy of the ejected electron is $K = E_{photon} - |E_1| = 15 \ eV - 13.6 \ eV = 1.4 \ eV$.$5$. Therefore,the detector will observe one photon of $10.2 \ eV$ and one electron of $1.4 \ eV$.

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