Animated Solution for Physics - Atoms and Nuclei: If the atom ${}_{100}{\text{Fm}}^{257}$ follows the Bohr's model and the radius of last orbit of ${}_{100}{\text{Fm}}^{257}$ is $n$ times the Bohr radius, then find $n$

Imagine you are looking at a massive atom of Fermium (${}_{100}{\text{Fm}}^{257}$). The question asks us to assume that this multi-electron atom strictly follows the Bohr model. Our goal is to find the radius of its outermost orbit and express it as a multiple of the Bohr radius ($a_0$).

To do this, we first need to determine which orbit is the outermost one. In the Bohr-Bury scheme, the maximum number of electrons that can be accommodated in the $m^{\text{th}}$ shell is given by the formula $2m^2$.

Let's start filling the 100 electrons of Fermium into the shells:

- For $m=1$, capacity = $2(1{)}^2=2$ electrons. - For $m=2$, capacity = $2(2{)}^2=8$ electrons. - For $m=3$, capacity = $2(3{)}^2=18$ electrons. - For $m=4$, capacity = $2(4{)}^2=32$ electrons.

If we sum up the electrons in the first four shells, we get:

Fermium has a total of 100 electrons. The remaining electrons are:

These 40 electrons will jump into the next available shell, which is the 5th shell ($m=5$). Therefore, the outermost orbit for this atom is $m=5$.

Now, recall the formula for the radius of the $m^{\text{th}}$ orbit in a hydrogen-like atom according to the Bohr model:

where $Z$ is the atomic number and $a_0$ is the Bohr radius ($0.529\text{ A˚}$).

We know that for Fermium, the atomic number $Z=100$, and we just found that the outermost orbit is $m=5$. Let's substitute these values into our radius formula:

Simplifying the fraction, we get:

The question states that the radius is $n$ times the Bohr radius ($r_5=n{a}_0$). Comparing the two expressions, we can clearly see that:

Isn't it fascinating? Despite being in the 5th orbit, the massive nuclear charge of $+100e$ pulls the electrons in so tightly that the orbit's radius is actually just one-fourth of the radius of a hydrogen atom's first orbit!