Question

the force between an Alpha particle and electron separated by a distance of 1 angstrom in vacuum​

Answer 1

electrostatic force F_e = \dfrac{1}{4 \pi \epsilon_0} \dfrac{4e \times e}{({1 \times 10^{-10})}^2}F

e

=

4πϵ

0

1

(1×10

−10

)

2

4e×e

now switch the values

e = 1.602 × 10^(-19) C

\epsilon_0ϵ

0

= 8.85 × 10^(-12) C/V-m

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Answer 2

Charge of alpha particle = 2×e

Coulumbic force = k q(e) q(alpha) /r^2

=( 9×10^9 × 2 × (1.6×10^-19)^2 ) / (10^-10)^2

= 46.08 ×10^-9 N