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Atoms and Nuclei question

2016 · Q137
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Atoms and Nuclei question

2016 · Q137

NEETPhysicsAtoms and NucleiMCQ+4 / −1
When an α\alphaα-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as
  1. A
    1m2{1 \over {{m^2}}}m21​
  2. B
    m
  3. C
    1m{1 \over m}m1​
  4. D
    1m{1 \over {\sqrt m }}m​1​
View written solutionFree

Correct answer: C

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.

Kinetic energy = 12mv2{1 \over 2}m{v^2}21​mv2

Potential energy = KQqr{{KQq} \over r}rKQq​

∴\therefore∴ 12mv2{1 \over 2}m{v^2}21​mv2 = KQqr{{KQq} \over r}rKQq​

⇒\Rightarrow⇒ r ∝\propto∝ 1m{1 \over m}m1​

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