Question #114041

Through what angle will a 5-MeV alpha particle be scattered as it approaches a gold nucleus with an impact parameter of 2.6x10-13m?(Given that Z for Gold=79)


Determine the impact parameter of a 8-MeV alpha particle scattered by 12.5o if it approaches a gold nucleus.

Expert's answer

1)

b=kZe2cot⁡0.5θKb=\frac{kZe^2\cot{0.5\theta}}{K}

2.6⋅10−13=(9⋅109)(79)(1.6⋅10−19)2cot⁡0.5θ5⋅1.6⋅10−132.6\cdot 10^{-13}=\frac{(9 \cdot 10^{9})(79)(1.6\cdot 10^{-19})^2\cot{0.5\theta}}{5\cdot 1.6\cdot 10^{-13}}

θ=10°\theta=10\degree

2)


b=(9⋅109)(79)(1.6⋅10−19)2cot⁡258⋅1.6⋅10−13b=\frac{(9 \cdot 10^{9})(79)(1.6\cdot 10^{-19})^2\cot{25}}{8\cdot 1.6\cdot 10^{-13}}

b=3.05⋅10−14mb=3.05\cdot 10^{-14}m


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