Respuesta :
Answer:
The fraction of particles is [tex]6.21 \times 10^{-4}[/tex]
Explanation:
Given :
Alpha particle atomic no. [tex]Z_{1} = 2[/tex]
Gold foil atomic number [tex]Z_{2} = 79[/tex]
Thickness [tex]t = 10^{-8}[/tex] m
Density [tex]\rho = 19.3[/tex] [tex]\frac{g}{cm^{3} }[/tex]
Kinetic energy of alpha particle [tex]K = 5 \times 1.6 \times 10^{-19} \times 10^{6}[/tex] V
[tex]K = 8 \times 10^{-13}[/tex] V
Scattered angle [tex]\alpha =[/tex] 4.5°
Fraction of incident particle scattered at angle [tex]\alpha[/tex] is,
[tex]f(\alpha ) = \pi nt (\frac{Z_{1}e Z_{2} e }{8\pi \epsilon _{o} K } )^{2} \cot ^{2} (\frac{\alpha }{2} )[/tex]
Where [tex]n =[/tex] number density of particle, [tex]\epsilon _{o} = 8.85 \times 10^{-12}[/tex]
For calculating value of [tex]n[/tex]
[tex]n = \frac{\rho N_{A} }{M}[/tex]
Where [tex]N_{A} = 6.022 \times 10^{23}[/tex], [tex]M = 197[/tex] ( Mass number of gold foil )
[tex]n = 5.9 \times 10^{28}[/tex] [tex]\frac{atom}{m^{3} }[/tex]
Put the all value in above equation,
[tex]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} )[/tex]
[tex]f (4.5 ) = 0.96 \times 647.78 \times 10^{-6}[/tex]
[tex]f (4.5) = 6.21 \times 10^{-4}[/tex] [tex]\frac{atoms}{m^{2} }[/tex]
Therefore, the fraction of particles is [tex]6.21 \times 10^{-4}[/tex]
