Question

A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the electron is a distance of 7.5 × 10−9 m from the gold nucleus. (a) What is the magnitude of the electric force exerted by the gold nucleus on the electron?

Answer 1

Answer:

$F=3.2345*10^{-10}N$

Explanation:

Given data

Distance r=7.5×10⁻⁹m

Charge of electron -e= -1.6×10⁻¹⁹C

Charge of proton e=1.6×10⁻¹⁹C

To find

Electric force F

Solution

From Coulombs law we know that:

$F=K\frac{q_{1}q_{2} }{r^{2} }$

q₁ is charge of electron

q₂ is the charge of gold nucleus which contains 79 positively charge protons and 118 neutral neutrons.  

The Charge of single proton e=1.6×10⁻¹⁹C

79 proton charge q₂=79×1.6×10⁻¹⁹=1.264×10⁻¹⁷C

So

$F=\frac{1}{4\pi *8.85*10^{-12} } \frac{-1.6*10^{-19}*1.264*10^{-17}}{(7.5*10^{-9})^{2} }\\ F=3.2345*10^{-10}N$