9.1.Find the force of attraction betweenthenucleus of a hydrogen
atom and an electron. The radius of the hydrogen atom isO.5x10-8cm
and the charge of the nucleus is equal in magnitude and opposite in
sign to that of the electron.
Solution
Charge of electron
Q"=1.6\\times10^{-19}C"
Radius
R=O.5x10-8 cm
Then electrostatic force between electron and nucleus
"F=\\frac{KQQ}{R^2}\\\\=\\frac{9\\times10^9\\times1.6\\times10^{-19}\\times 1.6\\times10^{-19} }{( 0.5\\times10^{-10}) \n ^2}\\\\=92.16\\times10^{-9}N."
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