A hydrogen atom consists of one proton and one electron. In the simplest model f

Question

A hydrogen atom consists of one proton and one electron. In the simplest model for the hydrogen atom, the electron orbits the proton in a circular path of radius r = 5.29 Times 10^-11 m. Calculate the magnitude of the electric force between the proton and electron when they are that far apart. Use your answer from part (a) to calculate the magnitude of the (centripetal) acceleration of the electron caused by the unbalanced electric force. Using the fact that the centripetal acceleration is a_c = v^2/r, calculate the speed of the orbiting, electron. Use your answer from part (c) to calculate the electron's kinetic energy.

Explanation / Answer

The charges on an electron and a proton are -1.6*10-19 C and 1.6*10-19 C; the radius of separation is r = 5.29*10-11 m.

a) The magnitude of the electric force between the two is given by

FE = (1/40)*(q1q2/r2) where 0 = 8.85*10-12 C2/N.m2 is the permittivity of free space and q1 and q2 denote the charges. Therefore,

FE = [1/4*3.14*(8.85*10-12 C2/N.m2)]*(-1.6*10-19C)*(1.6*10-19 C).(5.29*10-11 m)2 = -8.229*10-8 N = -8.23*10-8 N (ans).

b) This electric force must be balanced by the centripetal force acting on the electron. The centripetal force acts opposite to the electric force such that

FC = -FE.

Also, FC = m.ac2 where ac = centripetal acceleration of the electron and m = 9.11*10-31 kg; thus,

(9.31*10-31 kg)*ac2 = 8.23*10-8 N

====> ac2 = 8.23*10-8 N/(9.31*10-31 kg) = 8.8399*1022 m2/s4

====> ac = 2.973*1011 m/s2 2.97*1011 m/s2 (ans).

c) Given ac = v2/r where v = velocity of the electron, we have,

2.97*1011 m/s2 = v2/(5.29*10-11 m)

===> v2 = 15.7113 m2/s2

===> v = 3.963 m/s 3.96 m/s (ans).

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.