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- Homework Statement
- For a hydrogen atom, the normalised wavefunction of an electron in the 1s state, assuming a point nucleus, is $$\psi (r) = \left( \frac {1}{\pi a_0^3} \right)^{1/2} e^{-r/a_0}$$, where ##a_0## is the Bohr radius. Find an approximate expression for the probability of finding the electron in a small sphere of radius ##r_b << a_0## centered on the proton. What is the electronic charge density?

- Relevant Equations
- None.

Hi. I would love if someone could check my solution since me and the answer sheet I found online don't agree.

The probability is given by the triple integral

\begin{align*}

\int_0^{r_b} \int_0^{2\pi} \int_0^\pi |\psi (r)|^2 r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= \frac{1}{\pi a_0^3}\int_0^{r_b} \int_0^{2\pi} \int_0^\pi e^{-2r/a_0} r^2 \sin{\theta} \,d\theta \,d\phi \,dr\\

&= \frac{4}{a_0^3} \int_0^{r_b} r^2 e^{-2r/a_0} \,dr \\

&= \frac{4}{a_0^3} \left( -\frac{a_0}{2} e^{-2r/a_0} \left( r^2 +a_0r +\frac{a_0^2}{2} \right) \right) \Bigg|_0^{r_b} \\

&= 1 - \frac{2}{a_0^2} e^{-2r_b/a_0} \left( r_b^2 +a_0 r_b + \frac{a_0^2}{2} \right) \\

&\approx 1 - e^{-2r_b/a_0}

\end{align*}

As for the electron charge density, ##\rho##, I figure that since

\begin{align*}

\int_0^{r_b} \int_0^{2\pi} \int_0^\pi \rho r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= Q

\end{align*}

where ##Q## is the total charge, then ##\rho## must be equal to ##e|\psi|^2##, where ##e## is the elementary charge.

But according to the answer sheet I found online, the probability is given by

\begin{align*}

\int_0^{r_b} |\psi|^2 r^2 \,dr \approx \frac{r_b^3}{\pi a_0^3} e^{-2r_b/a_0} \sim \left( \frac{r_b}{a_0} \right)^3

\end{align*}

I find the answer sheet dubious, since it only integrates over the radius and skips the angular part. In addition, I am unable to see how it solves the integral to get that solution, but I've been wrong many times before. Anyone who can provide input?

Edit: Accidentally posted mid-edit, sorry!

The probability is given by the triple integral

\begin{align*}

\int_0^{r_b} \int_0^{2\pi} \int_0^\pi |\psi (r)|^2 r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= \frac{1}{\pi a_0^3}\int_0^{r_b} \int_0^{2\pi} \int_0^\pi e^{-2r/a_0} r^2 \sin{\theta} \,d\theta \,d\phi \,dr\\

&= \frac{4}{a_0^3} \int_0^{r_b} r^2 e^{-2r/a_0} \,dr \\

&= \frac{4}{a_0^3} \left( -\frac{a_0}{2} e^{-2r/a_0} \left( r^2 +a_0r +\frac{a_0^2}{2} \right) \right) \Bigg|_0^{r_b} \\

&= 1 - \frac{2}{a_0^2} e^{-2r_b/a_0} \left( r_b^2 +a_0 r_b + \frac{a_0^2}{2} \right) \\

&\approx 1 - e^{-2r_b/a_0}

\end{align*}

As for the electron charge density, ##\rho##, I figure that since

\begin{align*}

\int_0^{r_b} \int_0^{2\pi} \int_0^\pi \rho r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= Q

\end{align*}

where ##Q## is the total charge, then ##\rho## must be equal to ##e|\psi|^2##, where ##e## is the elementary charge.

But according to the answer sheet I found online, the probability is given by

\begin{align*}

\int_0^{r_b} |\psi|^2 r^2 \,dr \approx \frac{r_b^3}{\pi a_0^3} e^{-2r_b/a_0} \sim \left( \frac{r_b}{a_0} \right)^3

\end{align*}

I find the answer sheet dubious, since it only integrates over the radius and skips the angular part. In addition, I am unable to see how it solves the integral to get that solution, but I've been wrong many times before. Anyone who can provide input?

Edit: Accidentally posted mid-edit, sorry!

Last edited: