In a Hydrogen Atom in the Unexcited State, the Probability $F(x)=\frac{4}{\left(a_{0}\right)^{3}} \int_{0}^{x} r^{2} e^{\frac{-2 r}{a_{0}}} d r$

Multiple Choice

In a hydrogen atom in the unexcited state, the probability of finding the sole electron within x meters of the nucleus is given by $F(x) =\frac{4}{\left(a_{0}\right) ^{3}} \int_{0}^{x} r^{2} e^{\frac{-2 r}{a_{0}}} d r$ , for $x \geq 0$ , where $a_{0}=5.29 \times 10^{-11}$ meters.What is its corresponding probability density function f(x) ?

A) $f(x) =\frac{4}{\left(a_{0}\right) ^{3}} \cdot 2 x \cdot \frac{-2}{a_{0}} e^{\frac{-2 x}{a_{0}}}$
B) $f(x) =\frac{4}{\left(\alpha_{0}\right) ^{3}} \cdot\left(\frac{-2}{a_{0}} x^{2} e^{\frac{-2 x}{a_{0}}}+2 x e^{\frac{-2 x}{a_{0}}}\right)$
C) $f(x) =\frac{4}{\left(a_{0}\right) ^{3}} x^{2} e^{\frac{-2 x}{a_{0}}}$
D) $f(x) =-\frac{12}{\left(a_{0}\right) ^{4}} x^{2} e^{\frac{-2 x}{a_{0}}}$