Normalising simplified proton distribution

[alfredbester]

I tried asking this question in the maths help, but am still stuck.

Q. For a simplified model of a proton's charge distribution, (where R can be considered as some characteristic "size" of the proton):$$\rho$$(r) $$\propto$$ $$(1/r)$$Exp(- r / R) where R is some characteristic size of the proton.

With some help I've figured the integral to this solution is the exponentialintegral: Ei(-r/R) in the past when I'm asked to normalise something I'd take the limits from infinity to -infinity and find N so the probability is 1, here though the exponential integral can't be normalised from what I can see. Any pointers would be appreciated.

 

[Meir Achuz]

d^3r in spherical coords with no angular dependence is 4 pi r^2dr so the integral rho r^2 dr converges.

 

[alfredbester]

So

$$\rho$$(r) = (4pi r^2 dr).(1/r)Exp(-r / R)

Integral from infinity to 0 of |$$\rho$$(r)|^2 = N^2 . [(1/r^2).Exp(-2r / R)]

is 1 = N^2. (4pi) [1 / (-2/R)] = N^2 .2piR => N = 1 / sqrt(2.pi.R)