Solveeit Logo
Login

Question

Physics Question on Nuclear physics

In an alpha particle scattering experiment, distance of closest approach for the α\alpha particle is 4.5×10−14 m4.5 \times 10^{-14} \, \text{m}. If the target nucleus has atomic number 80, then maximum velocity of α\alpha-particle is ______ ×105 m/s\times 10^5 \, \text{m/s} approximately.
(14πϵ0=9×109 SI unit, mass of α particle=6.72×10−27 kg)\left( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{SI unit}, \, \text{mass of } \alpha \, \text{particle} = 6.72 \times 10^{-27} \, \text{kg} \right)

Answer

The distance of closest approach is given by:
rmin=4KZe2mv2.r_{\text{min}} = \frac{4KZe^2}{mv^2}.
Rearranging for velocity:
v=4KZe2mrmin.v = \sqrt{\frac{4KZe^2}{mr_{\text{min}}}}.
Substitute values:
v=4⋅9⋅109⋅80⋅(1.6×10−19)26.72×10−27⋅4.5×10−14.v = \sqrt{\frac{4 \cdot 9 \cdot 10^9 \cdot 80 \cdot (1.6 \times 10^{-19})^2}{6.72 \times 10^{-27} \cdot 4.5 \times 10^{-14}}}.
Simplify:
v=156×105 m/s.v = 156 \times 10^5 \, \text{m/s}.
Final Answer:
156×105 m/s156 \times 10^5 \, \text{m/s}.

Explanation

Solution

The distance of closest approach is given by:
rmin=4KZe2mv2.r_{\text{min}} = \frac{4KZe^2}{mv^2}.
Rearranging for velocity:
v=4KZe2mrmin.v = \sqrt{\frac{4KZe^2}{mr_{\text{min}}}}.
Substitute values:
v=4⋅9⋅109⋅80⋅(1.6×10−19)26.72×10−27⋅4.5×10−14.v = \sqrt{\frac{4 \cdot 9 \cdot 10^9 \cdot 80 \cdot (1.6 \times 10^{-19})^2}{6.72 \times 10^{-27} \cdot 4.5 \times 10^{-14}}}.
Simplify:
v=156×105 m/s.v = 156 \times 10^5 \, \text{m/s}.
Final Answer:
156×105 m/s156 \times 10^5 \, \text{m/s}.