Answer
(a) $6.4\times 10^6m/s$
(b) $0.27pm$
(c) less than
Work Step by Step
(a) We can determine the speed of the $\alpha$ particle as follows:
$K=\frac{1}{2}m_{\alpha}v^2$
$\implies v^2=\frac{2K}{m_{\alpha}}$
We plug in the known values to obtain:
$v^2=\frac{2(0.85MeV)}{4(931.5MeV/c^2)}$
$v^2=4.563\times 10^{-4}c^2$
$\implies v=\sqrt{4.563\times 10^{-4}(3\times 10^8)^2}$
$v=6.4\times 10^6m/s$
(b) We know that when the alpha particle approaches the closest distance, then the speed becomes zero and according to the law of conservation of energy, all the initial kinetic energy is converted into potential energy.
We know that
$K_i=0.85MeV=(0.85MeV)(1.6\times 10^{-13}J/MeV)$
$K_i=1.36\times 10^{-13}J$
Now, according to the conservation of energy
$U_f=K_i$
$\implies \frac{2(Ze)(2e)}{d}=K_i$
This simplifies to:
$d=\frac{2(9\times 10^9Nm^2/C^2)(79)(1.6\times 10^{-19}C)^2}{(1.36\times 10^{-13}J)}$
$d=2.6767\times 10^{-13}m$
$d=0.27pm$
(c) As the atomic mass of copper is less than that of the gold and the distance of closest approach is directly proportional to atomic mass, hence the distance of closest approach for the copper nucleus is less than that of gold.
