A proton carries a charge of 1.60X10 -19 C and has a radius of 8.42X10 -16 m. To

Question

A proton carries a charge of 1.60X10-19 C and has a radius of 8.42X10-16 m. To get two protons to fuse, they must be brought so close together that the strong nuclear attraction overcomes the Coulomb repulsion--they must be touching.

*Calculate the force with which the two protons must repel each other when they are just touching & Calculate what the initial acceleration of each proton would be while noting that the force is huge especially when we consider that the mass of the proton is only 1.67X10-27 kg.

Explanation / Answer

Required Force F = k*q1*q2/r^2

k = 9*10^9

q1 = q2 =1.6*10^-19 C

r = disatnce between them = 2*8.42*10^-16 = 16.84*10^-16 m

F = k*q1*q2/r^2 = (9*10^9*1.6*10^-19*1.6*10^-19)/(2*8.41*10^-16)^2

F = 81.43 N

here F is the Force acting on each proton by another proton

but according to newton's second law of motion

Fnet = m*a

F = m*a

81.43 = m*a

here m is the mass of the proton = 1.67*10^-27 kg

81.43 = (1.67*10^-27*a)

a is the accelaration of each proton

a = (81.43)/(1.67*10^-27)

a = 4.87*10^28 m/s^2

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