Question

A chemist has a piece of foil that is approximately \(5 \times 10^{4}\) atoms thick. If an alpha particle must come within \(10^{-12} \mathrm{~cm}\) of a nucleus for deflection to occur, what is the probability that an alpha particle will be deflected, assuming the nuclei are not directly behind one another? Assume that the area of one atom is \(4 \times 10^{-16} \mathrm{~cm}^{2}\).

Short answer

The probability that an alpha particle will be deflected is approximately \(P(\text{deflection}) \approx \frac{\pi}{4} \times 10^{-8}\).

Step-by-step solution

Find the area in which deflection occurs

To find the area where deflection occurs, we can consider the nucleus as a circle with a radius of \(10^{-12}\) cm (since deflection occurs when the alpha particle comes within \(10^{-12}\) cm of the nucleus). Then, we can calculate the area using the formula for the area of a circle, which is \(A = \pi r^{2}\), where A is the area, and r is the radius. In this case, \(r = 10^{-12} \mathrm{~cm}\), so the area of deflection is: \[A = \pi (10^{-12})^{2} = \pi \times 10^{-24} \mathrm{~cm}^{2}\]

Find the ratio of deflection area to total area

Now, we'll find the ratio of the area in which deflection occurs (A) to the total area of one atom. The total area of one atom is given as \(4 \times 10^{-16} \mathrm{~cm}^{2}\). We can calculate the ratio as follows: \[\text{Ratio} = \frac{\pi \times 10^{-24} \mathrm{~cm}^{2}}{4 \times 10^{-16} \mathrm{~cm}^{2}}\]

Simplify the ratio and find the probability of deflection

Now, let's simplify the ratio and find the probability of deflection. \[\text{Ratio} = \frac{\pi \times 10^{-24}}{4 \times 10^{-16}} = \frac{\pi}{4} \times \frac{10^{-24}}{10^{-16}} = \frac{\pi}{4} \times 10^{-8}\] Since the ratio obtained represents the probability of an alpha particle being deflected, the probability of deflection is approximately: \[P(\text{deflection}) \approx \frac{\pi}{4} \times 10^{-8}\]

Key concepts

Probability

Understanding probability in this chemistry problem involves calculating the likelihood that an alpha particle will be deflected when it comes near a nucleus. In simple terms, probability measures how often a specific event might happen. The step-by-step solution calculates the probability of deflection by comparing the area where deflection occurs to the total area of one atom. This gives a ratio, which serves as the probability.
- The area calculated for deflection is the space within which, should an alpha particle enter, deflection will happen.
- This tiny region is then measured against the area of an entire atom, showing just how likely or unlikely deflection is given random motion.
In this case, the solution shows that the probability of deflection is expressed as \(P(\text{deflection}) \approx \frac{\pi}{4} \times 10^{-8}\). This small number indicates that an alpha particle has a very low chance of hitting a nucleus directly for deflection.

Nuclear Physics

Nuclear physics is the field that studies atomic nuclei and their constituents and interactions. In this exercise, we deal with alpha particles and their interactions with atomic nuclei. - Alpha particles are composed of two protons and two neutrons. They are positively charged and come from radioactive decay processes.
- When alpha particles approach a nucleus, they can be deflected due to the electromagnetic forces between the positively charged particle and the nucleus.
Understanding deflections in nuclear physics helps in the study of atomic structure, nuclear reactions, and even applications in radiation therapy.
In our specific scenario, examining these deflections helps understand the forces at play within a seemingly solid piece of material and gives insights into atom arrangements.

Atom Structure

The structure of an atom is fundamental to understanding how alpha particles might be deflected. An atom consists of a nucleus surrounded by electrons moving in a cloud. - The nucleus is incredibly small compared to the entire atom, often only taking up a tiny fraction of its volume.
- The nucleus is where most of an atom's mass resides, made up of protons and neutrons.
- In scenarios such as ours, where alpha particles interact with atoms, it is the nucleus that matters most because of its concentrated positive charge.
Alpha particles, due to their size and charge, are primarily affected by the charge of the nucleus. Therefore, understanding that deflections occur within an exceedingly small space aids in grasping the high degree of precision needed in chemical calculations and radiation physics.

Chemistry Calculations

In the context of this exercise, chemistry calculations involve determining exact probabilities and understanding reactions at the atomic level. - Calculations first involve finding areas where interactions can occur, such as the circular area around the nucleus for potential deflections.
- These calculations require formulas from geometry, like the area of a circle \(A = \pi r^2\), with the radius being the critical distance where interaction is possible.
- By comparing this specific area to the overall area of an atom, we estimate probabilities that help predict the behavior of matter on a small scale.
Such calculations are central to both predicting outcomes in experiments and designing situations like radiation shielding, where understanding interactions on a small scale can have massive implications.