A) 10.2eV
B) 20.4eV
C) 12.1eV
D) 16.8eV
Correct Answer: B
Solution :
Let v = speed of neutron before collision \[\frac{1}{\sqrt{3}}\frac{\tau }{Bi}\]speed of neutron after collision \[{{m}_{1}}\]speed of hydrogen atom after collision and \[{{m}_{2}}\] energy of the excitation From the conservation of linear momentum, \[{{m}_{1}}\] ...(i) From the conservation of energy, \[{{m}_{2}}\] ...(ii) from Eq. (i),\[{{m}_{1}}\] From Eq. (ii),\[d\sqrt{\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}}\] \[d\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}}\]\[d\sqrt{\frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}}\] \[d\sqrt{\frac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}}\]\[=20m{{s}^{-1}}\] \[(\text{take}\,g=10\text{ }m{{s}^{-}}^{2})\] As,\[I={{I}_{0}}\sin \omega t,\]must be real\[{{I}_{0}}=10\]\[\omega =100\pi \,\text{rad/s}\] or\[\pi \] The maximum energy that can be absorbed by hydrogen atom in ground state to go into excited state is 10.2 eV. Therefore, 'the minimum kinetic energy required is \[\pi \]
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