Question

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated towards a gold nucleus and its path was substantially deflected by the Coulomb interaction. If the initial kinetic energy of the doubly charged alpha nucleus was 5.76 MeV, how close to the gold nucleus (79 protons) could it come before being turned around

Answer 1

Answer:

$3.95 \times 10^{-14}m$

Explanation:

We are given that

Charged on alpha particle=q=2e=$2\times 1.6\times 10^{-19} C$

Where $e=1.6\times 10^{-19} C$

Initial kinetic energy=K.E=5.76 MeV=$5.76\times 10^6\times 1.6\times 10^{-9} C$

$1 Me V=10^6\times 1.6\times 10^{-19} V$

Z=79

Charge on protons=$q'=79\times 1.6\times 10^{-19} C$

We have to find the closeness of alpha particle to the gold nucleus before being turned around.

Initial kinetic energy=Final potential energy

$5.76\times 10^6\times 1.6\times 10^{-9}=\frac{Kq_1q_2}{r}$

Where $k=9\times 10^9$

$r=\frac{9\times 10^9\times 79\times 1.6\times 10^{-19}\times 2\times 1.6\times 10^{-19}}{5.76\times 10^6\times 1.6\times 10^{-9}}$

$r=3.95\times 10^{-14}m$