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

• agnimusayoti
A result is the result of an integration. In this case, there is no integration because you are solving for a variable that is not given.f

#### 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

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.

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

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

• etotheipi
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.

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.

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

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