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Chapter 7: Problem 21

A uranium nucleus (mass 238 u), initially at rest, undergoes radioactive decay. After an alpha particle (mass \(4.0 \mathrm{u}\) ) is emitted, the remaining nucleus is thorium (mass \(234 \mathrm{u}) .\) If the alpha particle is moving at 0.050 times the speed of light, what is the recoil speed of the thorium nucleus? (Note: "u" is a unit of mass; it is not necessary to convert it to kg.)

Short Answer

Expert verified
Answer: The recoil speed of the thorium nucleus is approximately -8.55 x 10^{-4} times the speed of light, indicating a motion in the opposite direction to the emitted alpha particle.

Step by step solution

01

Understand the conservation of linear momentum

In any collision or interaction, the total linear momentum of a closed system remains constant. In this case, the uranium nucleus undergoes radioactive decay which involves the emission of the alpha particle. Therefore, we have a closed system. Initially, the uranium nucleus is at rest, so the total linear momentum of the system is zero. By the conservation of linear momentum, the total linear momentum should remain zero after the alpha particle is emitted. We can use this information to find the recoil speed of the thorium nucleus.
02

Set up the linear momentum equation

Since the total linear momentum is conserved, we can write the equation for the initial and final linear momenta of the system. Let \(v_\text{Th}\) be the recoil speed of the thorium nucleus. Initially, linear momentum = 0 Finally, linear momentum = \((m_\alpha v_\alpha + m_\text{Th} v_\text{Th})\) Now, since linear momentum is conserved, \(0 = m_\alpha v_\alpha + m_\text{Th} v_\text{Th}\)
03

Solve for the recoil speed of the thorium nucleus

We can rewrite the equation to solve for \(v_\text{Th}\): \(v_\text{Th} = -\frac{m_\alpha v_\alpha}{m_\text{Th}}\) We are given: - Mass of alpha particle \(m_\alpha = 4.0 \ \text{u}\) - Speed of alpha particle \(v_\alpha = 0.05 \ c\) where \(c\) is the speed of light - Mass of thorium nucleus \(m_\text{Th} = 234 \ \text{u}\) Plug in these values into the equation to find the recoil speed of the thorium nucleus: \(v_\text{Th} = -\frac{4.0 \ \text{u} \times 0.05 \ c}{234 \ \text{u}}\) Notice that the mass units (\text{u}) cancel out, and we are left with: \(v_\text{Th} = -\frac{0.2 \ c}{234}\)
04

Calculate the result

Finally, calculate the recoil speed of the thorium nucleus: \(v_\text{Th} \approx -8.55 \times 10^{-4} c\) Thus, the recoil speed of the thorium nucleus is approximately -8.55 x 10^{-4} times the speed of light. The negative sign indicates that the thorium nucleus recoils in the opposite direction to the motion of the alpha particle.

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Most popular questions from this chapter

For a system of three particles moving along a line, an observer in a laboratory measures the following masses and velocities. What is the velocity of the \(\mathrm{CM}\) of the system? $$\begin{array}{cc}\hline \text { Mass }(\mathrm{kg}) & v_{x}(\mathrm{m} / \mathrm{s}) \\\\\hline 3.0 & +290 \\\5.0 & -120 \\\2.0 & +52 \\\\\hline\end{array}$$
A radioactive nucleus is at rest when it spontaneously decays by emitting an electron and neutrino. The momentum of the electron is $8.20 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ and it is directed at right angles to that of the neutrino. The neutrino's momentum has magnitude $5.00 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ (a) In what direction does the newly formed (daughter) nucleus recoil? (b) What is its momentum?
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