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What is the distance of closest approach...

What is the distance of closest approach of an alpha particle to a silver nucleus is the kinetic energy of the alpha particle is 0.40 MeV.

Text Solution

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The correct Answer is:
`3.4xx10^(-13)m`
The approach distance of an alpha-particle to the nucleus will be a minimum in the case of a central collision, when the entire kinetic energy of the alpha-particle is transformed into potential energy: `K=U=(Z_(1)Z_(2)e^(2))/(4piepsi_(0)r),"where "Z_(1)andZ_(2)` are the respective atomic numbers.
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The distance of closest approach of an alpha particle to the nucleus of gold is 2.73 xx 10^(-14) m. Calculate the energy of the alpha particle.

Write the expression for distance of closest approach for a alpha -particle.

Knowledge Check

  • The distance of the closest approach of an alpha particle fired at a nucleus with kinetic of an alpha particle fired at a nucleus with kinetic energy K is r_(0) . The distance of the closest approach when the alpha particle is fired at the same nucleus with kinetic energy 2K will be

    A
    `4 r_(0)`
    B
    `r_(0)/2`
    C
    `r_(0)/4`
    D
    `2r_(0)`

  • A stationary radioactive nucleus of mass 210 units disintegrates into an alpha particle of mass 4 units and residual nucleus of mass 206 units. If the kinetic energy of the alpha particle is E, then the kinetic energy of the residual nucleus is

    A
    `((2)/(105))E`
    B
    `((2)/(103))E`
    C
    `((103)/(105))E`
    D
    `((103)/(2))E`

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