Question

An alpha particle having kinetic energy 10Mev.an atom having atomic mass 35.calculate its distance of closest approach

Answer 1

Answer:

10.08 fermi

Explanation:

At it's closest approach, $Kinetic~Energy=Potential~Energy$

$Kinetic~Energy=10~MeV=1.6\,\times 10^{-12}\\\\Potential~Energy=K\times \frac{q_1\,q_2}{r}$

 where, q₁ = charge on alpha particle = 2 × 1.6 × 10⁻¹⁹

              q₂ = charge on nucleus = 35 × 1.6 × 10⁻¹⁹

              r = distance of closest approach

              K = 1/4π∈₀

Hence,

∴$K.E.=K\times \frac{q_1q_1}{r}\\\\r=\frac{K\times q_1q_2}{K.E.}\\\\r=\frac{9\times 10^9\times 2\times 1.6\times 10^{-19}\times 35\times 1.6\times 10^{-19}}{1.6\times 10^{-12}}\\\\r=1.008\times 10^{-14}~meters\\\\r=10.08~fermi$