Answer:
2.12 x 10^-4 N
Explanation:
charge on uranium nucleus, Q = 92 e
charge on proton, q = e
distance, d = 1 x 10^-11 m
The force between the two charged particles is given by
e = 1.6 x 10^-19 C
F = 2.12 x 10^-4 N
Thus, the force between the proton and the nucleus of Uranium is 2.12 x 10^-4 N.
The answer is approximately 2.998e+8
Sorry I honestly don’t know that answer
Answer: F = 3.28 × 10^-7 N
Explanation:
Alpha particle charge (q)= 2e
e = 1.6 × 10^-19
q = 2e = 2 × (1.6 *10^-19) = 3.2 × 10^-19
1/4πEo = 9 × 10^9
Distance(r) = 5.3 × 10^-11m
The force of attraction between the two particles is given by:
F = (1/4πEo) (q1q2 / r^2)
F = [(9 × 10^9) (3.2 × 10^-19) (3.2 × 10^-19)] / (5.3 × 10^-11)^2
F = (92.16 × 10^(9 - 19 - 19)) / 28.09 × 10^-22
F = (92.16 × 10^-29) / 28.09 × 10^-22
F = 3.28 × 10^(-29 + 22)
F = 3.28 × 10^-7 N