Animated Solution for Chemistry - Atomic Structure: Which one of the following about an electron occupying the $1s$-orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by $a_0$)

Imagine you are trying to locate a hyperactive firefly in a completely dark room. You can't predict its exact path, but if you take a million photographs and overlay them, a glowing cloud emerges. This is exactly how we visualize the electron in the quantum mechanical model of the hydrogen atom. The electron doesn't orbit the nucleus in neat, planetary circles; instead, it exists as a probability cloud defined by its wave function, $\psi$.

In this problem, we are tasked with identifying the incorrect statement regarding an electron in the $1s$-orbital. To do this, we must dissect the subtle differences between probability density, radial probability, and the energy of quantum states.

Radial Probability

Let's address the concept of probability density, denoted by $|\psi {|}^2$ or $R^2(r)$. This value tells us the probability of finding the electron in a tiny, specific unit of volume at a distance $r$ from the nucleus. For any $s$-orbital, the wave function is spherically symmetric and has no angular nodes. Mathematically, the function $R(r)$ is an exponential decay curve that is at its absolute maximum right at the nucleus ($r=0$). Therefore, the probability density is indeed maximum at the nucleus.

However, space is three-dimensional. If we want to know the probability of finding the electron at a distance $r$, regardless of direction, we must consider the volume of a thin spherical shell at that distance. The volume of this shell is $4\pi r^{2}dr$.

When we multiply the probability density by the volume of the shell, we get the radial probability distribution function:

At the nucleus ($r=0$), the volume of the shell is zero, so $P(0)=0$. As we move away from the nucleus, the $r^2$ term grows while the $R^2(r)$ term shrinks. The competition between these two terms creates a peak. For the $1s$-orbital, this peak occurs exactly at $r=a_0$, the Bohr radius. This means $a_0$ is the most probable distance to find the electron.

Furthermore, the tail of this distribution curve extends to infinity. It never truly hits zero. This means there is a non-zero probability of finding the electron at $2a_0$, $3a_0$, or even a mile away (though infinitesimally small).

What about the energy of the electron? The Virial Theorem is a profound principle in mechanics that relates the average kinetic energy to the average potential energy for stable, bound systems. For a system bound by an inverse-square force (like the Coulombic electrostatic force between a proton and an electron), the theorem states:

Because kinetic energy is always positive, the potential energy is negative (indicating a bound state). If we take the magnitude, we see that $|⟨PE⟩|=2⟨KE⟩$. The magnitude of the potential energy is exactly double the kinetic energy on average.

Finally, we arrive at the crux of the problem: the total energy of the electron. In quantum mechanics, an electron in a specific orbital (like the $1s$-orbital) is in a stationary state. This means its total energy is a constant, quantized value. For the hydrogen atom, the energy of the $n$-th state is given by:

For the $1s$ state ($n=1,Z=1$), the total energy is strictly $-13.6\text{ eV}$. It does not fluctuate or reach a "maximum" as the electron moves closer to or further from the nucleus. The distance $a_0$ is simply the location of maximum radial probability, not maximum energy.

Therefore, the statement claiming that the total energy is maximum at $a_0$ is fundamentally flawed, making it the correct answer to our question.