Respuesta :
Answer:
[tex]F=3.2345*10^{-10}N[/tex]
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:
[tex]F=K\frac{q_{1}q_{2} }{r^{2} }[/tex]
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
[tex]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[/tex]
The magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].
What is electric force?
Electric force is the force of attraction or repulsion between two charged particles. It can be given as,
[tex]F=K\dfrac{q_1\times q_2}{r^2}[/tex]
Here [tex]k[/tex] is coulomb's constant, [tex]q[/tex] is charge on the objects and [tex]r[/tex] is the distance between two objects.
Given information-
The number of proton in gold atom is 79.
The number of neutrons in gold atom is 118.
The distance of the electron from the nucleus is [tex]7.5 \times 10^{-9} \rm m[/tex].[tex]r[/tex]
- a) The magnitude of the electric force exerted by the gold nucleus on the electron-
The charge on electron is [tex]-1.6\times 10^{-19} C[/tex] and the charge on the proton is [tex]1.6\times 10^{-19} C[/tex].
Put the values in the above equation as,
[tex]F=8.98\times10^9\times\dfrac{(79\times1.6\times10^{-19})\times1.6\times10^{-19}}{(7.5\times10^{-19})^2}\\F=3.235\times10^{-10}\rm N[/tex]
Hence the magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].
Learn more about the electric force here;
https://brainly.com/question/14372859
