Question

An atom is modelled as a point charge of +e at the nucleus with the electron charge distribution, described by the charge density

$\rho =\frac{-15e}{8\pi a^3}(1-\frac{r^2}{a^2})$ for $r\le a<0$

$\rho \simeq 0$ for $r>a,$ where 'a' is a constant.

The net charge contained within a sphere of radius r < a is :

• A. $[\frac{3r^5}{2a^5}-\frac{4r^4}{3a^4}]e$
• B. $[\frac{3r^5}{2a^5}-\frac{5r^3}{2a^3}]e$
• C. $[\frac{3r^5}{2a^5}-\frac{5r^3}{2a^3}+1]e$
• D. $\text{Z}ero$

Answer 1

Answer: (C)

$[\frac{3r^5}{2a^5}-\frac{5r^3}{2a^3}+1]e$

Net charge within the sphere of radius $r\le a$
$e+{\int }_0^r\left(4\pi r^{2}dr\right)\rho =e+4\pi \left(\frac{-15e}{8\pi a^3}\right){\int }_0^r(1-\frac{r^2}{a^2})r^{2}dr$
$=e(1+\frac{3r^5}{2a^5}-\frac{5r^3}{2a^3})$