Problem 25.30 The nucleus of a atom (an isotope of the element xenon with mass 1

Question

Problem 25.30 The nucleus of a atom (an isotope of the element xenon with mass 125 ) is 6.0 in diameter. It has 54 protons and charge . Part A What is the electric force on a proton 2.0 from the surface of the nucleus? Hint: Treat the spherical nucleus as a point charge. Express your answer to two significant figures and include the appropriate units. = SubmitMy AnswersGive Up Part B What is the proton's acceleration? Express your answer to two significant figures and include the appropriate units. = SubmitMy AnswersGive Up

Explanation / Answer

Solution:-

the total charge on nucleus
q=54*1.6*10^-19 C
q=8.64*10^-18 C

A).
F=kqQ/r^2
here Q=charge on proton=1.6*10^-19 C
r=[(6/2)+2]*10^-15 m=5*10^-15 m
thus
F=9*10^9 * 8.64*10^-18 *1.6 *10^-19/(5*10^-15)^2
F=497.664 N


B).mass of proton=1.67*10^-27 kg

F=ma
497.664=1.67*10^-27 *a

a=2.98*10^29 m/s^2

Alternate:

The total distance between a nucleus as a point charge and the proton is:

r = (6.0 x 10^-15 m) + (2.0 x 10^-15 m) = 8.0 x 10^-15 m

The charge of the nucleus is:

Q = 54 x (1.6 x 10^-19 C) = 8.64 x 10^-18 C

The electric force on the proton is:

F = kQq / r²

= ((9.0 x 10^9 Nm²/C²) x (8.64 x 10^-18 C) x (1.60 x 10^-19 C)) / (8.0 x 10^-15 m)²

= 194.4 N

The proton's acceleration is:

a = F / m = 194.4 N / (1.67 x 10^-27 kg) = 1.2 x 10^29 m/s²

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