© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the total charge Q as an integral
The total charge Q is obtained by integrating the volume charge density
$ \rho(r) = \frac{A}{r^2} e^{-\frac{2r}{a}} $ throughout the volume of the sphere of radius R.
In spherical coordinates, the volume element is $ dV = 4\pi r^2\,dr $, so
$$
Q \;=\; \int_{0}^{R} \rho(r)\,4\pi r^2\,dr \;=\;\int_{0}^{R} \frac{A}{r^2} e^{-\frac{2r}{a}}\,4\pi r^2\,dr.
$$
Step 2: Simplify the integrand
Notice that $ r^2 $ in the denominator cancels with $ r^2 $ in the volume element:
$$
Q \;=\; 4\pi A \int_{0}^{R} e^{-\frac{2r}{a}}\,dr.
$$
Step 3: Evaluate the integral
We integrate $ e^{-\frac{2r}{a}} $ with respect to r:
$$
\int_{0}^{R} e^{-\frac{2r}{a}}\,dr
\;=\; \left[ -\frac{a}{2} e^{-\frac{2r}{a}} \right]_{0}^{R}
\;=\; -\frac{a}{2}\Bigl(e^{-\frac{2R}{a}} - e^0\Bigr)
\;=\; \frac{a}{2} \Bigl(1 - e^{-\frac{2R}{a}}\Bigr).
$$
Hence,
$$
Q \;=\; 4\pi A \cdot \frac{a}{2}\,\Bigl(1 - e^{-\frac{2R}{a}}\Bigr)
\;=\; 2\pi a A\,\Bigl(1 - e^{-\frac{2R}{a}}\Bigr).
$$
Step 4: Solve for R
From
$$
Q \;=\; 2\pi a A \Bigl(1 - e^{-\frac{2R}{a}}\Bigr),
$$
isolate the exponential term:
$$
1 - e^{-\frac{2R}{a}} \;=\; \frac{Q}{2\pi a A},
$$
$$
e^{-\frac{2R}{a}} \;=\; 1 \;-\; \frac{Q}{2\pi a A}.
$$
Take the natural logarithm on both sides:
$$
-\frac{2R}{a} \;=\; \ln\!\Bigl(1 - \frac{Q}{2\pi a A}\Bigr).
$$
Therefore,
$$
R
\;=\; -\frac{a}{2} \,\ln\!\Bigl(1 - \frac{Q}{2\pi a A}\Bigr)
\;=\; \frac{a}{2} \,\ln\!\Bigl(\frac{1}{1 - \frac{Q}{2\pi a A}}\Bigr).
$$
Step 5: Final expression for the radius R
Hence, the radius R is
$$
\boxed{R = \frac{a}{2}\,\ln\!\Bigl(\frac{1}{1 - \frac{Q}{2\pi a A}}\Bigr).}
$$
This matches the given correct option.
