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MAVERICK [17]
2 years ago
15

Exercise 15.1 (a) Find the magnitude of the electric force between two protons separated by 1 femtometer (10^-15), approximately

the distance between two protons in the nucleus of a helium atom. (b) If the protons were not held together by the strong nuclear force, what would be their initial acceleration due to the electric force between them
Physics
1 answer:
Goryan [66]2 years ago
7 0

Answer:

(a) 230.4 N

(b) a = 1.38\times10^{29} m/s^{2}

Explanation:

Charge on a proton, q = 1.6\times10^{-19}C

Distance between the two protons, d = 1 femto meter = 1\times10^{-15}m

(a) By use of coulomb's law, the force between the two protons is given by

F =\frac{k\times q\times q}{d^{2}}

Where, k is the Coulombic constant = 9x 10^9 Nm^2/C^2

So, the force between them is given by

F =\frac{9\times10^{9}\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{\left (1\times10^{-15}  \right )^{2}}

F = 230.4 N

(b) Acceleration is the ratio of force to the mass of proton.

the mass of proton, m = 1.67 x 10^-27 kg

So, acceleration

a=\frac{230.4}{1.67\times10^{-27}}

a = 1.38\times10^{29} m/s^{2}

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