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A 1 MeV proton is sent against a gold le...

A 1 MeV proton is sent against a gold leaf `(Z=79)`. Calculate the distance of closest approach for head-on collision.

Text Solution

Verified by Experts

The correct Answer is:
`1.137xx10^(-13)m`
`d=(Ze^(2))/(4piepsi_(0)(1//2 mv^(2)))`.
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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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Knowledge Check

  • A beam of beryllium nucleus (z = 4) of kinetic energy 5.3 MeV is headed towards the nucleus of gold atom (Z = 79). What is the distance of closest approach?

    A
    `10.32xx10^(-14)m`
    B
    `8.58xx10^(-14)m`
    C
    `3.56xx10^(-14)m`
    D
    `1.25xx10^(-14)m`

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