What keeps the electrons from leaving an atom in the rutherford model of the atom?
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The magnitude of electric field is produced by the electrons at a certain distance.
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C
We have the equation of motion
, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the displacement.
In this case initial velocity = 2.20 m/s, final velocity = 0 m/s, displacement = 14 m
On substitution we will get 0 =
On solving we can find the acceleration value as -0.173
So free fall acceleration = 0.173