Question

find the electrostatic force of attraction between proton and electron in hydrogen atom ​

Answer 1

The force of electrostatic attraction is given by,

$\frac{1}{4\pi \: e_{0} } \frac{ q_{1} \: q_{2} }{ {r}^{2} }$

Where symbols hold their usual meaning,


To find r we must use the radius relation for 1st orbit,
$0.529{(1)}^{2} angstrom \\ \\ = 0.529 \: angstrom \\ \\ \\ = 5.29 \times {10}^{ - 11} m$


Using this value,

we have,

$- 9 \times {10}^{9} \times \frac{1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{ {(5.29 \times {10}^{ - 11} )}^{2} } N \\ = - 9 \times \frac{ {(1.6)}^{2} }{ {(52.9)}^{2} } \times {10}^{9 + ( - 38) + 22} \times N \\ = 0.823 \times {10}^{ - 7} \times N\\ \\ = 8.23 \times {10}^{ - 8} N$