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An alpha particle of velocity 1.6xx10^(7...

An alpha particle of velocity `1.6xx10^(7) ms^(-1)` approaches a gold nucleus (Z=79). Calculate the distance of the closest approach. Mass of an alpha particle is `6.6xx10^(-27)` kg. What is the significance of this closest approach?

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The correct Answer is:
`4.3 xx 10^(-14)m` , the size of the nucleus cannot be greater than `4.3 xx 10^(-14)m`.
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NN GHOSH-ELECTRIC FIELD AND POTENIAL-EXERCISE-B
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  2. Two equal positive point charges are sperated by a distance of 2a. A p...

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  3. An alpha particle of velocity 1.6xx10^(7) ms^(-1) approaches a gold nu...

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  6. Two equal charges, +Q each, are at a distance r from each other. A thi...

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  7. An infinite number of charges, each equal to q are arranged in a line ...

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  11. A thin copper ring of radius a is charged with q units of electricity....

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  12. A an electron is placed at the centre of the copper ring of problem 37...

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  14. A thin nonconducting ring of radius r has a linear charge density q=q0...

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  15. A thin wire ring of radius r has an electric charge q. What is the ten...

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  16. A charge +Q is uniformly distributed over a thin ring of a radius R. F...

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