Question

Question 5. Calculate the probability that an electron in the ground state of the Hydrogen
atom is outside the classically allowed region, i.e. (r > 2a0).

Answer 1

Answer on Question #64243-Physics-Quantum Mechanics

Calculate the probability that an electron in the ground state of the Hydrogen atom is outside the classically allowed region, i.e. $(r > 2a0)$.

Solution

The probability of finding an electron in the forbidden region is:

$$P = \int_{\text{forbidden region}} d^3 \boldsymbol{r} \, |\psi_{100}(\boldsymbol{r})|^2 = \int_{2a_0}^{\infty} dr r^2 \int d\Omega \, |\psi_{100}(\boldsymbol{r})|^2 = \int_{2a_0}^{\infty} dr r^2 \int d\Omega \left[ \frac{1}{\sqrt{4\pi}} (2) \left(\frac{1}{a_0}\right)^{\frac{3}{2}} e^{-\frac{r}{a_0}} \right]^2 = \frac{4}{a_0^3} \int_{2a_0}^{\infty} dr r^2 e^{-\frac{2r}{a_0}} = \frac{4}{a_0^3} \left[ e^{-\frac{2r}{a_0}} \left(-\frac{a_0 r^2}{2} - \frac{a_0^2 r}{2} - \frac{a_0^3}{4}\right) \right]_{2r_0}^{\infty} = \frac{4}{a_0^3} e^{-4} \left(2a_0^3 + a_0^3 + \frac{a_0^3}{4}\right) = 13e^{-4} \approx 0.24.$$

Answer: $13e^{-4} \approx 0.24$.

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