Question

A beam of some kind of practical of velocity $2\times {10}^{7}m/s$ is scattered by a gold (z=79) foil. Find specific charge of this particle charge/mass if the distance of closest approach is $7.9{10}^{-14}m$

Answer 1

Since particle is seathered so potential energy between charge of beam and charge of gold nucleus is equal to their kinetic energy

$\frac{a_1{a}_2}{r}=\frac{1}{2}m{v}^2$

$a_1=$ charge of gold nucleus

$a_2=$ charge of beam

$m=$ mass of beam

$r=$ distance of close approach

$v=$ velocity of bean

$k=$ constant $=9\times {10}^9$

$\frac{a_2}{m}=\frac{v^{2}r}{2k{a}_1}$

$\frac{a_2}{m}=\frac{(2\times {10}^7{)}^2\times (7.9\times {10}^{-14})}{2\times 9\times {10}^9\times (79\times 1.6\times {10}^{-19})}$

specific charge $=1.3\times {10}^8$