Radius of proton given radius of electron

[t_n_p]

Homework Statement

http://img518.imageshack.us/img518/9337/untitled2gr1.jpg

The Attempt at a Solution

(a) Using the right hand rule, my sketch looks like the following

http://img340.imageshack.us/img340/5530/untitled3mk0.jpg

(b) F=qvbsin(theta)
F=(1.6*10^-19)*(5*10^7)*(0.5)*sin(90deg)
F= 4*10^-12 Newtons

(c) Using F = (mv²)/r
4*10^-12 = [(9.1*10^-31)*(5*10^7)²]/r
r = 5.7*10^-4 m

(d) I used the formula [Rp/Re] = [MpVp]/[MeVe], but the final answer is simply a radius which is 2000 times that of the electron. For some reason, I don't think it's right!

 

[andrevdh]

(d)

$$F_e = F_p$$

therefore

$$\frac{m_e v^2}{r_e} = \frac{m_p v^2}{r_p}$$

giving

$$\frac{r_e}{m_e} = \frac{r_p}{m_p}$$

so that

$$\frac{m_p}{m_e} r_e = r_p$$

or

$$r_p = 2000\ r_e$$

 

[t_n_p]

hmm so I was right! :rofl: