Animated Solution for Chemistry - Atomic Structure: P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is $4\pi r^2\mathrm{dr}$. The qualitative sketch of the dependence of P on r is -

Imagine you are playing hide and seek with an electron inside a hydrogen atom. Classical physics told us the electron runs on a fixed track, like a planet around the sun. But quantum mechanics shatters this illusion. The electron is everywhere and nowhere until we look for it. Instead of a fixed orbit, we have a probability cloud.

This problem tests a very subtle but crucial distinction in quantum mechanics: the difference between the probability of finding an electron at a specific point versus finding it in a specific region.

When students first learn about the 1s orbital, they are introduced to the wave function, ${\psi }_{1s}(r)$. For a hydrogen atom, this function looks like:

If we want to know the probability density (the probability per unit volume) at a specific point, we square the wave function to get ${\psi }^2$. Because of the $e^{-2r/a_0}$ term, this density is actually highest right at the nucleus ($r=0$) and decays exponentially as we move away.

If the question had asked for the plot of probability density, Graph (D) would be the correct answer. But the question asks for something entirely different!

The question specifically asks for the probability $P$ of finding the electron in a spherical shell of infinitesimal thickness $dr$ at a distance $r$ from the nucleus.

Think of an onion. We aren't asking for the density of the onion at the center; we are asking how much onion material is in a specific layer. To find this, we must multiply the density by the volume of that layer. The volume of a thin spherical shell is the surface area of a sphere multiplied by its thickness:

Now, we construct our master equation for the radial probability $P(r)$:

This equation represents a beautiful mathematical tug-of-war.

1. At the nucleus ($r=0$): The $r^2$ term is exactly zero. Even though the density ${\psi }^2$ is maximum here, the volume of a shell with zero radius is zero. Therefore, $P(0)=0$. The graph must start at the origin.

2. Far away ($r\to \infty$): As we move outward, the $r^2$ term grows rapidly, trying to pull the probability up. However, the exponential term $e^{-2r/a_0}$ decays even faster. In the battle between polynomials and exponentials, the exponential always wins at infinity. Therefore, $P(\infty )\to 0$.

Since the function starts at zero, must be positive everywhere else, and eventually returns to zero, it must have a peak somewhere in between.

Looking at our options, only Graph (B) starts at the origin, rises to a maximum, and then decays asymptotically.

Bonus Insight: If you use calculus to differentiate $P(r)$ with respect to $r$ and set it to zero, you will find that the peak occurs exactly at $r=a_0$. This means the most probable distance to find the electron is exactly the Bohr radius! Quantum mechanics doesn't discard Bohr's model entirely; it simply reinterprets his fixed orbit as the peak of a probability distribution.