A) 2 photon of energy 10.2 eV
B) 2 photon of energy 1.4 eV
C) One photon of energy 10.2 eV and an electron of energy 1.4 Ev
D) One photon of energy 10.2 eV and another photon of energy 1.4 eV
Correct Answer: C
Solution :
Loss in Kinetic Energy during collision between moving proton and stationary hydrogen atom is used to excite electron from ground state to first excited state. Because \[{{E}_{2}}-{{E}_{1}}=10.2\,\,eV\] So, first photon will excite the hydrogen atom in ground state, thus, a photon of 10.2 eV will be released during de-excitation. The atom can be ionised by second photon of energy 15 eV. moving proton and stationary hydrogen Balance energy = energy retained by electron. = (15 - 13.6) = 1.4eV Thus, an electron of energy 1.4 eV will be released by second photon
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