What fraction of 5 MeV α particles will be scattered through angles greater than 4.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thi

Question

What fraction of 5 MeV α particles will be scattered through angles greater than 4.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thickness 10-8 m?

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Gerda 5 months 2021-08-10T23:46:26+00:00 1 Answers 7 views 0

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    2021-08-10T23:47:53+00:00

    Answer:

    The fraction of particles is 6.21 \times 10^{-4}

    Explanation:

    Given :

    Alpha particle atomic no. Z_{1} = 2

    Gold foil atomic number Z_{2} = 79

    Thickness t = 10^{-8} m

    Density \rho = 19.3 \frac{g}{cm^{3} }

    Kinetic energy of alpha particle K = 5 \times 1.6 \times 10^{-19} \times 10^{6} V

    K = 8 \times 10^{-13} V

    Scattered angle \alpha  = 4.5°

    Fraction of incident particle scattered at angle \alpha is,

       f(\alpha ) = \pi nt (\frac{Z_{1}e Z_{2} e  }{8\pi \epsilon _{o} K } )^{2} \cot ^{2} (\frac{\alpha }{2} )

    Where n = number density of particle, \epsilon _{o} = 8.85 \times 10^{-12}

    For calculating value of n

     n = \frac{\rho N_{A}  }{M}

    Where N_{A} =  6.022 \times 10^{23}, M = 197           ( Mass number of gold foil )

    n = 5.9 \times 10^{28} \frac{atom}{m^{3} }

    Put the all value in above equation,

    f(4.5 ) = 3.14 \times  5.9 \times 10^{28} \times 10^{-8}  (\frac{2 \times 79  (1.6 \times 10^{-19} ) ^{2}  }{8\times 3.14 \times 8.85 \times 10^{-12} 8 \times 10^{-13}  } )^{2} \cot ^{2} (\frac{4.8 }{2} )

    f (4.5 ) = 0.96 \times 647.78 \times 10^{-6}

    f (4.5) = 6.21 \times 10^{-4} \frac{atoms}{m^{2} }

    Therefore, the fraction of particles is 6.21 \times 10^{-4}

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