Answer
${\bf 24.7}\;\rm cm$
Work Step by Step
We know that each collision results in an energy loss of 30 eV.
So, the total number of collisions $N$ required to stop the alpha particle is given by
$$
N =\dfrac{E_i}{E_{\rm collision}} $$
Plug the known;
$$
N = \frac{5.0 \times 10^6}{30} = \bf 1.667 \times 10^5\;\rm collision
$$
To calculate the mean free path ($ \lambda $) for an alpha particle in air or nitrogen gas, we’ll use the formula:
$$
\lambda = \frac{1}{4 \sqrt{2 \pi} \left( \frac{N}{V} \right) r^2}
$$
But the root of this formula is considering the radius of the cylinder is $2r$ and here we need a radius of $r$. You can review chapter 18, the part of Mean Free Path.
So, this formula became;
$$\lambda = \frac{1}{ \sqrt{2 \pi} \left( \frac{N}{V} \right) r^2}\tag 1$$
where:
- $ \frac{N}{V} $ is the number density of molecules in the medium (air or nitrogen),
- $ r $ is the effective interaction radius of the alpha particle with the surrounding molecules.
Assume Standard Temperature and Pressure (STP):
At STP (1 atm pressure and 293 K), the number density of gas molecules (for an ideal gas) is approximately:
$$
\frac{N}{V} = \frac{p}{k_B T} = \frac{1.013\times 10^5 \, \text{Pa}}{(1.38 \times 10^{-23} \, \text{J/K})(273 \, \text{K})} =\bf 2.69 \times 10^{25} \, \rm{m}^{-3}
$$
We’ll use this value for air or nitrogen gas at room temperature, as they have similar properties.
An alpha particles have a small nuclear radius ($ 1.5 \times 10^{-15} \, \text{m} $), but their effective interaction radius in terms of ionization is much larger.
We can approximate an interaction radius of $ r \approx 10^{-10} \, \text{m} $ (a conservative assumption based on ionization effects, and also it is the same distances between nitrogen molecules since the alpha particle is too small compared to them).
Substitute Values into (1)
$$
\lambda = \frac{1}{ \sqrt{2 \pi} (2.69 \times 10^{25} )(10^{-10} )^2}
$$
$$
\lambda =\bf 1.483\times 10^{-6} \, \rm {m}
$$
The stopping distance is the total distance the particle will travel, which is the product of the mean free path and the number of collisions.
$$
d_{\rm Stopping} = N \lambda= (1.667 \times 10^5)(1.483\times 10^{-6} )
$$
$$
d_{\rm Stopping} \approx \color{red}{\bf 24.7}\;\rm cm$$
