Question

The wave function for a hydrogen atom in the 2s state is psi_2s® = 1/squareroot 32 pi a^3 (2-r/a) e^-r/2a. In the Bohr model, the distance between the electron and the nucleus in the n=2 state is exactly  Calculate the probability that an electron in the 2s state will be found at a distance less than 4a from the nucleus. P=

Answer 1

\(\psi_{2 s}(r)=\frac{1}{\sqrt{32 \pi a^{3}}}\left(2-\frac{r}{a}\right) e^{-r / 2 a}\)

wave function of a Hydrogen atom in the \(2 s\) state is

Probability that an electron in the \(2 s\) state is found between 0 and \(4 a\) is

\(P(0 < r < 4 a)=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{4 a} r^{2} \psi^{*}(r) \psi(r) \sin \theta d r d \theta d \phi\)

\(P(0 < r < 4 a)=\frac{1}{\left(\sqrt{32 \pi a^{3}}\right)^{2}} \int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{4 a} r^{2}\left(2-\frac{r}{a}\right)^{2}\left(e^{-r / 2 a}\right)^{2} \sin \theta d r d \theta d \phi\)

\(P(0

\(P(0 < r < 4 a)=\frac{1}{8 a^{3}} \int_{0}^{4 a} r^{2}\left(2-\frac{r}{a}\right)^{2} e^{-r / a} d r\)

\(P(0 < r < 4 a)=\frac{1}{8 a^{3}} \int_{0}^{4 a}\left(2 r-\frac{r^{2}}{a}\right)^{2} e^{-r / a} d r\)

substitute \(u=\frac{r}{a}\)

\(d u=\frac{d r}{a}\)

\(P(0 < r < 4 a)=\frac{a}{8 a^{3}} \int_{0}^{4}\left(2 a u-\frac{a^{2} u^{2}}{a}\right)^{2} e^{-u} d u\)

\(P(0 < r < 4 a)=\frac{1}{8} \int_{0}^{4}\left(4 u^{2}+u^{4}-4 u^{3}\right) e^{-u} d u\)

\(P(0< r <4 a)=\frac{1}{8}(1.4064)=0.1758\)