91Ó°ÊÓ

The nucleus of an atom has a mass of \(3.80 \times 10^{-25} \mathrm{~kg}\) and is at rest. The nucleus is radioactive and suddenly ejects a particle of mass \(6.6 \times 10^{-27} \mathrm{~kg}\) and speed \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\). Find the recoil speed of the nucleus that is left behind. The particle flies off in one direction, the nucleus recoils away in the opposite direction, and momentum is conserved. Take the direction of the ejected particle as positive. We are given, \(m_{n i}=3.80 \times 10^{-25} \mathrm{~kg}\), \(m_{\mathrm{p}}=6.6 \times 10^{-27} \mathrm{~kg}, m_{n f}=m_{n i}-m_{p}=3.73 \times 10^{-25} \mathrm{~kg}\), and \(v_{p f}=1.5 \times 10^{7} \mathrm{~m} / \mathrm{s} ;\) find the final speed of the nucleus, \(\boldsymbol{v}_{n f}\) Solving leads to $$ -v_{n f}=\frac{\left(6.6 \times 10^{-27} \mathrm{~kg}\right)\left(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right)}{3.73 \times 10^{-25} \mathrm{~kg}}=\frac{10.0 \times 10^{-20}}{3.73 \times 10^{-25}}=2.7 \times 10^{5} \mathrm{~m} / \mathrm{s} $$ The fact that this is negative tells us that the velocity vector of the nucleus points in the negative direction, opposite to the velocity of the particle, which we took to be positive.

Step by step solution

01

Understand the concept of conservation of momentum

In an isolated system, the total momentum before and after an event must be conserved. For this exercise, the initial momentum of the system is zero because the nucleus is initially at rest.

02

Identify known values and variables

The mass of the initial nucleus \(m_{ni}=3.80 \times 10^{-25} \mathrm{~kg}\), the mass of the ejected particle \(m_p=6.6 \times 10^{-27} \mathrm{~kg}\), and its speed \(v_{pf}=1.5 \times 10^7 \mathrm{~m/s}\). The mass of the final nucleus \(m_{nf}=m_{ni} - m_p = 3.73 \times 10^{-25} \mathrm{~kg}\). We need to find the final speed of the recoil nucleus \(v_{nf}\).

03

Apply the conservation of momentum equation

The total initial momentum is zero. The equation for conservation of momentum after the particle is ejected is: \[ m_{ni} \cdot 0 = m_{nf} \cdot (-v_{nf}) + m_p \cdot v_{pf} \] This simplifies to: \[ 0 = m_{nf} \cdot (-v_{nf}) + m_p \cdot v_{pf} \]

04

Solve for the final velocity of the nucleus

Rearranging the equation gives \[ -v_{nf} = \frac{m_p \cdot v_{pf}}{m_{nf}} \] Substitute the known values: \[ -v_{nf} = \frac{(6.6 \times 10^{-27} \mathrm{~kg}) \cdot (1.5 \times 10^7 \mathrm{~m/s})}{3.73 \times 10^{-25} \mathrm{~kg}} = \frac{9.9 \times 10^{-20}}{3.73 \times 10^{-25}} = 2.7 \times 10^5 \mathrm{~m/s} \]

05

Interpret the negative sign of the velocity

The negative sign indicates that the recoil direction of the nucleus is opposite to the direction we defined as positive, which is the direction of the ejected particle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay

Radioactive decay is a process where an unstable atomic nucleus loses energy by emitting radiation. During this process, an atom transforms into a different element or isotope. This can happen in several ways, such as alpha decay, beta decay, or gamma decay. In our exercise, the nucleus undergoes radioactive decay by ejecting a particle.

This ejected particle is a key feature of radioactive decay, representing the nucleus's transition towards stability. The action of losing a particle is spontaneous and driven by the internal forces within the nucleus. As we see in this problem, the ejected particle has a specific mass and velocity, which are crucial in applying the laws of physics, such as the conservation of momentum.

Recoil Speed

Recoil speed refers to the velocity of an object after it has reacted to a force or an event. In our example, once the particle is ejected from the atomic nucleus, the remaining nucleus experiences a recoil.

This concept is similar to the recoil felt when firing a gun. The ejection of the bullet pushes the gun backward in response to the forward motion of the bullet. Here, the ejected particle causes the nucleus to move in the opposite direction, producing a recoil speed. Understanding recoil speed is essential because it helps us calculate how objects move after such interactions.

• The direction of recoil is opposite to the direction of the ejected particle's motion.
• The magnitude of this speed can be calculated using the conservation of momentum.

Momentum Conservation Equation

The momentum conservation equation is a fundamental principle in physics. It states that within an isolated system, the total momentum remains constant, before and after an event. In simple terms, momentum lost by one object is gained by another.

For our nuclear decay example, before the particle is ejected, the system's momentum is zero because the nucleus is at rest. The conservation of momentum principle allows us to set up an equation where the initial momentum equals the final momentum:

• Initial momentum: zero (since the nucleus is initially stationary).
• Final momentum: the sum of the momentum of the ejected particle and the recoil momentum of the nucleus.
• The equation used is: \[ 0 = m_{nf} imes (-v_{nf}) + m_p imes v_{pf} \]

The negative sign in the equation indicates the opposite direction of nucleus's velocity relative to the particle. Solving this equation gives us the recoil speed of the nucleus. Understanding this equation is vital for problems involving interactions where no external forces act on the system.