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 required

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 required to excite the atom from the ground state to the first excited state is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.$2$. When the first photon of $10.2 \, eV$ hits the atom,it is absorbed,and the atom transitions to the first excited state. After a microsecond,the atom de-excites,emitting a photon of $10.2 \, eV$.$3$. When the second photon of $15 \, eV$ hits the atom (which is now back in the ground state),since $15 \, eV > 13.6 \, eV$ (the ionization energy),the atom undergoes ionization.$4$. 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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