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The distance oif closest approach of an alpha-particle fired towards a nucleus with momentum p is r. What will be the distance of closest approach when the momentum of alpha-particle is `2p`?A. `2r`B. `4r`C. `r//2`D. `r//4` |
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Answer» Correct Answer - D KE=PE or`p^(2)/(2m)=1/(4pivarepsilon_(0))cdot(q1q2)/rRightarrowrpropto1/p^(2)` Hence, momentum of alpha-particle will be the closest approach of `r//4`. |
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