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The kinetic energy of an alpha - particl...

The kinetic energy of an `alpha` - particle which flies out of the nucleus of a `Ra^(226)` atom in radioactive disintegration is 4.78 MeV. Find the total energy evolved during the escape of the `alpha`-particle

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The correct Answer is:
4.87 MeV
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Knowledge Check

  • The kinetic energy of alpha -particle emitted in the alpha -decay of ""_(88)Ra^(266) is [given, mass number of Ra = 222 u]

    A
    `5.201` Me V
    B
    `3.301` Me V
    C
    `6.023` Me V
    D
    `4.871` Me V

  • 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 nucleus X initially at rest, undergoes alpha decay according to the equation _Z^232Xrarr_90^AY+alpha What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle?

    A
    (a) `90/92`
    B
    (b) `228/232`
    C
    (c) `sqrt(228/232)`
    D
    (d) `1/2`

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