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 the probability that the electron will be found within a sphere of radius $a_{0}$ ? Round to 3 decimal places.

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The Answer of 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 the probability that the electron will be found within a sphere of radius $a_{0}$ ? Round to 3 decimal places.

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 $F(x)=\frac{4}{\left(a_{0}\right)^{3}} \int_{0}^{x} r^{2} e^{\frac{-2 r}{a_{0}}} d r$