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

• agnimusayoti
In summary: 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.
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

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

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
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...

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

vanhees71

1. What is the average value of (1/r) for a 1s electron in the Hydrogen atom?

The average value of (1/r) for a 1s electron in the Hydrogen atom is equal to 1.5 x 10^7 m^-1. This value represents the average distance between the electron and the nucleus in the 1s orbital.

2. How is the average value of (1/r) calculated for a 1s electron in the Hydrogen atom?

The average value of (1/r) is calculated by taking the integral of the radial probability density function for the 1s orbital, which is given by (4πr^2ψ^2), where ψ is the wave function.

3. Does the average value of (1/r) change for different orbitals in the Hydrogen atom?

Yes, the average value of (1/r) will change for different orbitals in the Hydrogen atom. This is because the shape and size of the orbitals are different, resulting in different probabilities for the electron to be found at different distances from the nucleus.

4. How does the average value of (1/r) for a 1s electron in the Hydrogen atom compare to other atoms?

The average value of (1/r) for a 1s electron in the Hydrogen atom is relatively high compared to other atoms. This is due to the fact that Hydrogen only has one electron, so it is more likely to be found closer to the nucleus compared to atoms with multiple electrons.

5. Can the average value of (1/r) for a 1s electron in the Hydrogen atom be experimentally measured?

Yes, the average value of (1/r) for a 1s electron in the Hydrogen atom can be experimentally measured using spectroscopy techniques. By analyzing the energy levels of the electron in the 1s orbital, the average distance between the electron and the nucleus can be determined, and therefore the average value of (1/r) can be calculated.

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