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) $f(x)=-\frac{12}{\left(a_{0}\right)^{4}} x^{2} e^{\frac{-2 x}{a_{0}}}$

Question

Quizplus · Accepted Answer

The probability density function $f(x)$ is the derivative of the cumulative distribution function $F(x)$. Differentiating $F(x)$ with respect to $x$ gives $f(x)=\frac{4}{\left(a_{0}\right)^{3}} x^{2} e^{\frac{-2 x}{a_{0}}}$, which matches option C.

In a Hydrogen Atom in the Unexcited State, the Probability F(x)=4(a0)3∫0xr2e−2ra0drF(x)=\frac{4}{\left(a_{0}\right)^{3}} \int_{0}^{x} r^{2} e^{\frac{-2 r}{a_{0}}} d rF(x)=(a0​)34​∫0x​r2ea0​−2r​dr

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