charge of the Xe nucleus is q1 = +54 e charge on the proton is q2 = +e The radiu

Question

charge of the Xe nucleus is q1 = +54 e charge on the proton is q2 = +e The radius of the nucleus is r = 6.0/2 = 3.0 fm By treating the nucleus as a point charge the distance between the proton and the nucleus becomes d = r+2.3 = 3+2.3 = 5.3 fm = 5.3 Times 10-15 m Hence the electrostatic repulsive force on proton is F = 1/4 pi epsilon 0 q1q2/d2 or F = (9 Times 109) 54 e (e)/(5.3 Times 10-15)2 F = (9 Times 109) 54(1.6 Times 10-19)2/(28.09 Times 10-30 F = 44.2919 Times 10 = 442.919 N The acceleration experienced by the proton is a = F/m = 442.919 N/1.67 Times 10-27 kg = 265.22 Times 1027 m/s2 or a = 2.65 Times 1029 m/s2

Explanation / Answer

mass of proton is a constant = 1.67* 10^-27kg

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