Alpha-particle scattering experiment

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What is the distance of closest approach(in x10^-14 m) when a 5.0 MeV proton approaches a gold nucleus?

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Solution

At the distance r₀ of closest approach,
KE of a proton = PE of proton and the gold nucleus
$K = \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \cdot \frac{Z e \cdot e}{r_{0}} [q_{1} = Z e, q_{2} = e]$
$\begin{array}{rcl} o r r_{0} & = & \frac{1}{4 π ε_{0}} \cdot \frac{Z e^{2}}{K} \\ B u t K & = & 5.0 MeV = 5.0 \times 1.6 \times 10^{- 13} J \end{array}$
$For gold, Z = 79$
$\begin{array}{rcl} ∴ r_{0} & = & \frac{9 \times 10^{9} \times 79 \times {(1.6 \times 10^{- 19})}^{2}}{5.0 \times 1.6 \times 10^{- 13}} \\ = & 2.28 \times 10^{- 14} m ≃ 2.3 \times 10^{- 14} m \end{array}$

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Alpha-particle scattering experiment

Remember concepts with our Masterclasses.

What is the distance of closest approach(in x10-14 m) when a 5.0 MeV proton approaches a gold nucleus?

At the distance r0 of closest approach,KE of a proton = PE of proton and the gold nucleusK=12mv2=14πε0⋅Ze⋅er0q1=Ze,q2=eor r0=14πε0⋅Ze2KBut K=5.0MeV=5.0×1.6×10−13J For gold, Z=79∴ r0=9×109×79×1.6×10−1925.0×1.6×10−13 =2.28×10−14m≃2.3×10−14m

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What is the distance of closest approach(in x10^-14 m) when a 5.0 MeV proton approaches a gold nucleus?