What is the distance of closet approach when a 5 mev proton approaches a gold nucleus
Answer (1)
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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