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In a head-on collision between and alpha...

In a head-on collision between and `alpha`-particle and a gold nucleus, the distance of closest approach is 4.13 fermi. Calculate the energy of the particle. (1 fermi `= 10^(-15) m`)

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The correct Answer is:
5.51 MeV
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In a head on collision between an alpha particle and gold nucleus (Z=79), the distance of closest approach is 39.5 fermi. Calculate the energy of alpha -particle.

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Knowledge Check

  • In Geiger marsden experiment in a head on collision between an alpha particles and a gold nucleus the minimum separation is 4xx10^(-14) m . the energy of alpha particle is

    A
    5.68meV
    B
    8meV
    C
    4.47meV
    D
    7.24meV

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