Animated Solution for Chemistry - Atomic Structure: The ground state energy of hydrogen atom is $-13.6\text{ eV}$. The energy of second excited state of ${\text{He}}^{+}$ ion in eV is

When we study the Bohr model, the hydrogen atom is our starting point. We learn that the energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the elegant formula $E_n=-\frac{13.6}{n^2}\text{ eV}$. But what happens when we strip an electron from a helium atom, leaving behind a ${\text{He}}^{+}$ ion?

This ion is "hydrogen-like" because it only has one electron orbiting the nucleus. However, the nucleus now has two protons instead of one. This extra positive charge pulls the electron much tighter, significantly altering its energy levels.

To account for the stronger nuclear pull in hydrogen-like species (like ${\text{He}}^{+}$, ${\text{Li}}^{2+}$, etc.), we must introduce the atomic number $Z$ into our energy equation. The generalized Bohr energy formula becomes:

Here, $Z$ represents the atomic number (the number of protons in the nucleus), and $n$ is the principal quantum number (the orbit number).

A classic trap in atomic structure problems is the terminology of "excited states." The lowest possible energy level, where the electron is most stable, is the ground state, corresponding to $n=1$.

When the electron absorbs energy and jumps to the next available level, it reaches the first excited state, which is $n=2$.

Following this logic, the second excited state corresponds to $n=3$. This is a crucial detail! Many students mistakenly use $n=2$ when they see the word "second," leading to an incorrect calculation.

Now, let's apply our knowledge to the specific problem at hand. We need the energy of the second excited state of a ${\text{He}}^{+}$ ion.

1. Identify $Z$: For helium, the atomic number is $Z=2$. 2. Identify $n$: As established, the second excited state means $n=3$.

Let's substitute these values into our master equation:

Now, we perform the arithmetic. Multiplying $-13.6$ by $4$ gives us $-54.4$. Dividing $-54.4$ by $9$ yields exactly:

What does this $-6.04\text{ eV}$ actually mean? The negative sign indicates that the electron is bound to the nucleus. To completely free this electron from the $n=3$ orbit of the ${\text{He}}^{+}$ ion (i.e., to ionize it from this state), you would need to supply exactly $+6.04\text{ eV}$ of energy.

Notice how this compares to the hydrogen atom. The energy of the $n=3$ state in hydrogen is $-1.51\text{ eV}$. Because the helium nucleus has a $+2$ charge, it holds onto the electron much more tightly, resulting in a deeper (more negative) energy well.