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Chapter 40: Problem 59

How close can a \(5.00-\mathrm{MeV}\) alpha particle get to a uranium- 238 nucleus, assuming the only interaction is Coulomb?

Short Answer

Expert verified
The closest distance an alpha particle with a kinetic energy of 5.00 MeV can approach a uranium-238 nucleus is approximately 7.71 x 10^(-14) meters.

Step by step solution

01

1. Identifying the given information and defining the variables

We are given: - The kinetic energy of the alpha particle: \(E_k = 5.00 \ \mathrm{MeV}\) - The alpha particle has charge \(q_\alpha = 2e\), where \(e = 1.60 \times 10^{-19} \ \mathrm{C}\) is the elementary charge. - The charge of the uranium-238 nucleus, which has atomic number \(Z = 92\), and hence charge \(q_U = Ze\). We want to find the closest distance of approach, \(r_{min}\), between the alpha particle and the uranium-238 nucleus.
02

2. Convert energy from MeV to Joules

First, we should convert the given kinetic energy of the alpha particle from MeV (Mega-electronvolt) to Joules: $$ E_k = 5.00 \ \mathrm{MeV} \cdot \frac{1.60 \times 10^{-13} \ \mathrm{J}}{1 \ \mathrm{MeV}} = 8.00 \times 10^{-13} \ \mathrm{J} $$
03

3. Using conservation of energy

At the closest approach, the kinetic energy of the alpha particle will be converted entirely into potential energy. Thus, we can set up an equation for the conservation of energy: $$ E_k = \frac{kq_\alpha q_U}{r_{min}} $$ where \(k = 8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2}\) is the electrostatic constant.
04

4. Find the closest distance of approach

We want to find \(r_{min}\), so we can rearrange the conservation of energy equation to solve for it: $$ r_{min} = \frac{kq_\alpha q_U}{E_k} $$ Now, we can substitute the known values into the equation and calculate the closest distance of approach: $$ r_{min} = \frac{(8.99 \times 10^9 \ \mathrm{N \cdot m^2/C^2})(2 \cdot 1.60 \times 10^{-19} \ \mathrm{C})(92 \cdot 1.60 \times 10^{-19} \ \mathrm{C})}{8.00 \times 10^{-13} \ \mathrm{J}} $$ Finally, after performing the calculation, we find the closest distance between the alpha particle and the uranium-238 nucleus: $$ r_{min} \approx 7.71 \times 10^{-14} \ \mathrm{m} $$

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Most popular questions from this chapter

\(^{8} \mathrm{Li}\) is an isotope that has a lifetime of less than one second. Its mass is \(8.022485 \mathrm{u} .\) Calculate its binding energy in \(\mathrm{MeV}\).

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Write down equations to describe the alpha decay of the following atoms: a) \(^{212} \mathrm{Rn}\) b) \({ }^{241} \mathrm{Am}\)

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