Saturday, April 4, 2015

1.3e - The size of the hydrogen atom.

The expectation value for the potential energy (U) was calculated to be,

$$
<V> = \frac{1}{ \pi a^3} \int_{V} \Psi^*(x) V(x) \Psi(x) dV
$$
where \(V(r) = \frac{-e^2}{4 \pi \epsilon_0} \frac{1}{r}\) and \(dV=r^2 sin(\phi) drd\theta d\phi\).

$$
<V> = \frac{1}{ \pi a^3} \frac{-e^2}{4 \pi \epsilon_0} \int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{\infty} \frac{ e^{\frac{-2r}{a}}}{r} r^2 sin(\phi) drd\theta d\phi
$$

$$
<V> = \frac{4 \pi}{ \pi a^3} \frac{-e^2}{4 \pi \epsilon_0} \int_{0}^{\infty} r e^{\frac{-2r}{a}} dr = \frac{4 \pi}{ \pi a^3} \frac{-e^2}{4 \pi \epsilon_0} \left(\frac{a^2}{4} \right)
$$

$$
<V> = \frac{-e^2}{4 \pi \epsilon_0} \frac{1}{a}
$$

and similarly, the expectation value for the kinetic energy (T) was calculated to be,



$$
<T> = \frac{-\hbar^2}{2m \pi a^3} \int_{V} \Psi^*(x) \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial}{\partial r} \Psi(x)\right) dV
$$

$$
\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial}{\partial r}\right) = \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{-r^2}{a} e^{\frac{-r}{a}} \right) = \frac{1}{r^2} \left[ \frac{-2r}{a} e^{\frac{-r}{a}} + \left(\frac{-r^2}{a}\right) \left(\frac{-1}{a} \right) e^{\frac{-r}{a}} \right]
$$

$$
<T> = \frac{-\hbar^2}{2m \pi a^3} \int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{\infty} \left( \frac{-2r}{a} + \frac{r^2}{a^2} \right) e^{\frac{-2r}{a}} dr
$$
$$
<T> = \left( \frac{-\hbar^2 4 \pi}{2m \pi a^3} \right) \left[ \left(\frac{-2}{a} \right) \left(\frac{a^2}{4} \right) + \left( \frac{1}{a^2} \right) \left( \frac{a^3}{4} \right) \right]
$$
$$
<T> = \frac{ \hbar^2}{2ma^2}
$$
The value of ``a" that provides the lowest energy came out to be,
$$
\frac{ \partial E}{\partial a} = \frac{ \partial T}{ \partial a} + \frac{ \partial U}{ \partial a} = \frac{- \hbar^2(4ma)}{4m^2a^2} + \frac{e^2(4 \pi \epsilon_0)}{(4 \pi \epsilon_0 a)^2}
$$
$$
e^2 (4 \pi \epsilon_0)(m^2a^2) = \hbar^2 (4ma)(4 \pi^2 {\epsilon_0}^2)
$$
$$
a = \frac{4 \hbar^2 \pi \epsilon_0}{e^2 m_e}

$$
Here I have included a plot of the kinetic energy (T), the potential energy (V) and the total energy (E) all as functions of "a". T is in blue, V is in black and E is in red. The x-axis represents values of a, while the y-axis represents values of energies (in eV).
Graph 1: All energies plotted as functions of a.

So what do we learn from this? What does this mean? I am looking forward to seeing everyone's comments, thoughts, questions and discussion here.   ZS

2 comments:

  1. Damn! I was planning to type up me 1.3(e) today and submit it to Zack. I shouldn't haven taken that long nap last night!

    Anyways, what I found interesting about the kinetic energy problem is that taking to account that I didn't use the correct dV, I ended up getting the same magnitude as you did with a minus sign.

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  2. In saying stationary I mean an eigenvector of the energy operator, suggesting that the probability density is independent of time. The quantum state has a single definite energy as opposed to say, a quantum superposition of different energieswave functions.

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