Animated Solution for Physics - Atoms and Nuclei: An electron in an excited state of ${\text{Li}}^{2+}$ ion has angular momentum $\frac{3h}{2\pi }$. The de Broglie wavelength of the electron in this state is $p\pi a_0$ (where $a_0$ is the Bohr radius). The value of $p$ is

The journey to solving this problem begins with understanding the state of the electron. We are given that the electron in the ${\text{Li}}^{2+}$ ion has an angular momentum of $\frac{3h}{2\pi }$.

According to Bohr's second postulate, the angular momentum $L$ of an electron in a stable orbit is quantized. It must be an integral multiple of $\frac{h}{2\pi }$.

Mathematically, this is expressed as:

By comparing the given angular momentum with Bohr's quantization condition, we can easily deduce the principal quantum number $n$:

This tells us that the electron is currently residing in the third orbit.

Now that we know the electron is in the $n=3$ state, our next step is to find the physical size of this orbit.

For any hydrogen-like species, the radius of the $n^{\text{th}}$ orbit is given by the formula:

Here, $a_0$ is the Bohr radius (the radius of the first orbit of a hydrogen atom), and $Z$ is the atomic number of the nucleus.

For a lithium ion (${\text{Li}}^{2+}$), the atomic number $Z=3$. Substituting $n=3$ and $Z=3$ into our radius formula, we get:

So, the radius of this specific orbit is exactly three times the Bohr radius.

To find the de Broglie wavelength, we turn to the beautiful connection between particle waves and Bohr orbits.

Louis de Broglie proposed that for an electron orbit to be stable, the electron's matter wave must form a standing wave around the nucleus. This means the total circumference of the orbit must perfectly accommodate an integer number of wavelengths.

This profound physical insight is written as:

Let's apply this to our third orbit. We know the circumference is $2\pi r_3$, and the number of waves is $n=3$.

Substituting the value of $r_3$ we found earlier:

Simplifying this equation, the $3$ cancels out on both sides:

We have successfully calculated the de Broglie wavelength of the electron to be $2\pi a_0$.

The problem states that the wavelength is given by the expression $p\pi a_0$.

By directly comparing our calculated result with the given expression:

It is crystal clear that the value of $p$ must be $2$.

Final Answer: The value of $p$ is $2$.