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Calculate the nearest distance of approach of an `alpha`-particle of energy 2.5 Me V being scattered by a gold nucleus (Z=79).

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Calculate distance of closet approach by an alpha -particle of KE=2.5 MeV being scattered by gold nucleus (Z=79) .

Large angle scattering of alpha- particle led Rutherford to the discovery of atomic nucleus. It is the tiny central core of every atom in which entire positive charge of the atom is concentrated. The distance of closet approach of alpha particle from the nucleus is r_(theta)=((Ze)(2e))/(4pi epsilon_(theta)((1)/(2)mv^(2))) This distance r_(theta) gave him the order of size of nucleus. Read the above passage and answer the followign question: (i) What is the distance of closet approach of an alpha particle of energy 7.7 MeV from gold nucleus (z=79)? (ii) What is the implication of this relation in day to day life?

Knowledge Check

  • The closest distance of approach of an alpha -particle travelling with a velocity V towards a stationary nucleus is d. For the closest distance to become d//3 towards a stationary nucleus of double the charge, the velocity of projection of the alpha -particle has to be.

    A
    6V
    B
    `sqrt6V`
    C
    `(V)/(56)`
    D
    `(3V)/(2)`

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