Question

Consider the ground state of the H atom. (a) At what radius does the probability of fi nding an electron in a small volume located at a point fall to 25 per cent of its maximum value? (b) At what radius does the radial distribution function have 25 per cent of its maximum value? (c) What is the most probable distance of an electron from the nucleus? Hint: Look for a maximum in the radial distribution function.

Answer 1

Answer 2

(a). Radius where probability density is 25% of max

The probability density $\mathrm{\mid }\psi {\mathrm{\mid }}^2$ is highest at $r=0$. We set it to 25% of its peak:

$$e^{-2r\mathrm{/}{a}_0}=0.25⟹-2r\mathrm{/}{a}_0=\ln (0.25)⟹r=\frac{a_0}{2}\ln (4)\approx 0.693a_0\mathrm{.}$$

Answer: At $0.693a_0$ (~36.7 pm).

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(b). Radius where radial distribution $P(r)$ is 25% of max

The radial distribution $P(r)=4\pi r^2{e}^{-2r\mathrm{/}{a}_0}$ peaks at $r=a_0$. We solve:

$${\left(\frac{r}{a_0}\right)}^2{e}^{-2r\mathrm{/}{a}_0+2}=\mathrm{0.25.}$$

Numerical solutions give two radii:

• Inner solution: ~$0.30a_0$ (close to the nucleus).

• Outer solution: ~$2.94a_0$ (far from the nucleus).

Answer: $0.30a_0$$0.30a_0$ or $2.94a_0$$2.94a_0$.

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(c). Most probable distance :

The peak of $P(r)$ occurs where its derivative is zero:

$$\frac{dP}{dr}=0⟹r=a_0\mathrm{.}$$

Answer: Exactly $a_0$ (~52.9 pm).

Summary :-

• (a) 25% probability density at $0.693a_0$.

• (b) 25% radial probability at $0.30a_0$ or $2.94a_0$.

• (c) Most likely distance: $a_0$.