InterviewSolution
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InterviewSolution
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A neutron of energy 1 MeV and mass `1.6 xx 10^(-27)` kg passes a proton at such a distance that the angular momentum of the neutron relative to the proton approximately equals `10^(-33) J s`. The distance of closest approach neglecting the interaction between particles siA. `0.44nm`B. `0.44mm`C. `0.44Å`D. `0.44fm` |
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Answer» Correct Answer - d If `d` is the distance of closet approach given, then the angular momentum `=mvd=10^(-33) J s` `E=(1)/(2) mv^(2) =1 MeV =1.6 xx 10^(-33) J` Momentum, `P=sqrt(2m_nE)=sqrt(2 xx 1.6xx10^(27) xx1.6 xx 10^(-13))` `=1.6 sqrt2 xx 10 ^(20) kg m s^(1)` Distance of clossest approach, `d=(10^(33))/(1.6sqrt2xx10^(-20))` `=(1)/(1.6sqrt2) xx 10^(-13)=(100)/(1.6sqrt2)fm =0.44 fm` |
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