How far would the electron of a hydrogen atom have to be fromthe nucleus for the

Question

How far would the electron of a hydrogen atom have to be fromthe nucleus for the force of attraction to equal the weight of theelectron at the surface of the earth? I worked it out this way but got the wronganswer, what am I doing wrong? 8.99 x 109*((1.6 x10-19*1.6x10-19)/r2) = 9.11 x1031 How far would the electron of a hydrogen atom have to be fromthe nucleus for the force of attraction to equal the weight of theelectron at the surface of the earth? I worked it out this way but got the wronganswer, what am I doing wrong? 8.99 x 109*((1.6 x10-19*1.6x10-19)/r2) = 9.11 x1031 8.99 x 109*((1.6 x10-19*1.6x10-19)/r2) = 9.11 x1031

Explanation / Answer

     The charge of the electron andproton q1 = q2 = 1.60*10-19 C        The mass of electron ism1=9.1*10-31 kg      --------------------------------------------------------------            The force on the electrondue to proton is F = k*q1*q2 / r^2         But this forceis equal to the wegiht of the electron F = m1*g                                                                          So            k*q1*q2/ r^2 = m1*g                                              r = ( k*q1*q2 / m1*g )1/2                                                 = [ (8.99*109N.m2/C2)(1.60*10-19C)(1.60*10-19 C) / (9.1*10-31kg)*9.8m/s2 ]1/2                                                  =50.8 m                                                    =50.8 m  

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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