Verify the average value of (1/r) for a 1s electron in the Hydrogen atom

[agnimusayoti]

Homework Statement

Verify the average value of (1/r) for a 1s electron in the Hydrogen atom

Relevant Equations

For 1 electron, the wave function (n = 1, l = 0, ml = 0):
$$\Psi_{100}=\frac{e^{\frac{-r}{a_0}}}{(a_0)^{\frac{3}{2}}\sqrt{\pi}}$$
Average value of (1/r) therefore:
$$<1/r> = \int_{0}^{\infty} \frac{1}{r} |\Psi|^{2} dV$$

For spherical coordinate, $dV = r^{2} \sin {\theta} dr d\theta d\phi$
Therefore,
$$<\frac{1}{r}> = \frac{1}{(\pi)(a_0)^{3}} \int_{0}^{\infty} {r e^{\frac{-2r}{a_0}} dr \int_0^{\pi} \sin {\theta}} d\theta \int_0^{2\pi} d\phi$$
From partial integral, I've found:
$$\int_{0}^{\infty} r e^{-2r/{a_{0}}} dr = -\frac{a_0}{2} [r e^{-2r/a_0} + \frac{a_0}{2}]_{0}^{\infty}$$
Which I couldn't get $a_0^2/4$ as the result. Could you please show the mistake? Thankss

 

[LCSphysicist]

Homework Statement:: Verify the average value of (1/r) for a 1s electron in the Hydrogen atom
Relevant Equations:: For 1 electron, the wave function (n = 1, l = 0, ml = 0):
$$\Psi_{100}=\frac{e^{\frac{-r}{a_0}}}{(a_0)^{\frac{3}{2}}\sqrt{\pi}}$$
Average value of (1/r) therefore:
$$<1/r> = \int_{0}^{\infty} \frac{1}{r} |\Psi|^{2} dV$$

For spherical coordinate, $dV = r^{2} \sin {\theta} dr d\theta d\phi$
Therefore,
$$<\frac{1}{r}> = \frac{1}{(\pi)(a_0)^{3}} \int_{0}^{\infty} {r e^{\frac{-2r}{a_0}} dr \int_0^{\pi} \sin {\theta}} d\theta \int_0^{2\pi} d\phi$$
From partial integral, I've found:
$$\int_{0}^{\infty} r e^{-2r/{a_{0}}} dr = -\frac{a_0}{2} [r e^{-2r/a_0} + \frac{a_0}{2}]_{0}^{\infty}$$
Which I couldn't get $a_0^2/4$ as the result. Could you please show the mistake? Thankss

$$[r e^{-2r/a_0}]_{0}^{\infty} = 0 - 0 = 0$$ Ignore this term. Anyway, there is a wrong sign in your integration.

 

[vanhees71]

From partial integral, I've found:
$$\int_{0}^{\infty} r e^{-2r/{a_{0}}} dr = -\frac{a_0}{2} [r e^{-2r/a_0} + \frac{a_0}{2}]_{0}^{\infty}$$
Which I couldn't get $a_0^2/4$ as the result. Could you please show the mistake? Thankss

Check your integral!

Hint: An easy way to get it is to use Feynman's famous trick of "differentiating under the intgral sign"
$$\int \mathrm{d} r r \exp(-\lambda r) =-\frac{\mathrm{d}}{\mathrm{d} \lambda} \int \mathrm{d} r \exp(-\lambda r)=\ldots$$

 

[vela]

From partial integral, I've found:
$$\int_{0}^{\infty} r e^{-2r/{a_{0}}} dr = -\frac{a_0}{2} [r e^{-2r/a_0} + \frac{a_0}{2}]_{0}^{\infty}$$
Which I couldn't get $a_0^2/4$ as the result. Could you please show the mistake? Thankss

If you change variables from $r \to r'=r/(a_0/2)$, you end up with an integral which you hopefully recognize as the gamma function.

 

[Vanadium 50]

I couldn't get a₀²/4 as the result.

I should hope not. That has dimensions of area. The thing you are looking for has dimensions of inverse length.

 

[vanhees71]

The integral must have the dimension of area. The prefactor makes the dimensions right...

 

[Vanadium 50]

Ah, I was confused by the term "result".