- #1

agnimusayoti

- 240

- 23

- 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

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