Question

the probability of finding 1s electron in hydrogen atom in a given volume element dv is f(r)=4πr^2 A^2 e^-2r/a. determine the value of constant A and hence obtain the mean distance of electron from the origin .

Answer 1

$dV=4\pi r^2dr$

$4a^2\pi\int^{\infin}_0r^2e^{-2r/a}dr=1$

$4a^2\pi\cdot(-ae^{-2r/a}/4)(a^2+2ar+2r^2)|^{\infin}_0=1$

$\pi a^5=1$

$a=(\pi)^{1/5}=0.6$

$E(r)=4a^2\pi\int^{\infin}_0r^3e^{-2r/a}dr=4a^2\pi(-ae^{-2r/a}/8)(3a^3+6a^2r+6ar^2+4r^3)|^{\infin}_0=$

$=12a^6\pi /8=0.22$