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

AI Thread Summary
The discussion focuses on verifying the average value of (1/r) for a 1s electron in the Hydrogen atom using spherical coordinates. A specific integral, $$\int_{0}^{\infty} r e^{-2r/{a_{0}}} dr$$, is evaluated, but the user encounters difficulties in obtaining the expected result of $$a_0^2/4$$. Participants suggest checking the integration process and consider using techniques like differentiating under the integral sign to resolve the issue. They also clarify that the dimensions of the result must align correctly, emphasizing the importance of dimensional analysis in the calculations. The conversation highlights the complexities involved in quantum mechanical integrals and the need for careful mathematical handling.
agnimusayoti
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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
 
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agnimusayoti said:
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.
 
agnimusayoti said:
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"
$$\int \mathrm{d} r r \exp(-\lambda r) =-\frac{\mathrm{d}}{\mathrm{d} \lambda} \int \mathrm{d} r \exp(-\lambda r)=\ldots$$
 
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agnimusayoti said:
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.
 
agnimusayoti said:
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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Ah, I was confused by the term "result".
 
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