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What is the distance of closet approach when a 5 mev proton approaches a gold nucleus


Roor mal Sharma 12th Dec, 2020
Answer (1)
roopalsengar04 12th Dec, 2020

Dear,

when a particle is projected towards a nucleus, the condition for closest distance of approach is-

K.E of particle = P.E of system of particle and nucleus under closest distance

Therefore,

K.E of particle = 1/4πε o * qQ/d

5 MeV = 1/4πε o * (e)*(79e)/d

5x10 6 x1.6x10 -19 = 9x10 9 x [(1.6x10 -19 x79x1.6x10 -19 )/ d]

d = 23* 10 -15

d = 23 dermi



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