/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The nucleus of \(^{214} \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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The nucleus of \(^{214} \mathrm{Po}\) decays radioactively by emitting an alpha particle (mass \(6.65 \times 10^{-27} \mathrm{kg}\) ) with kinetic energy \(1.23 \times 10^{-12} \mathrm{J},\) as measured in the laboratory reference frame Assuming that the Po was initially at rest in this frame, find the recoil velocity of the nuclens that remains after the decay.

Short Answer

Expert verified
The recoil velocity of the nucleus is approximately \( -2.37 \times 10^4 \, \text{m/s} \).

Step by step solution

01

Identify the conservation principle

In radioactive decay, momentum is conserved. Before the decay, the initial momentum is zero since the Po nucleus is stationary. After emitting an alpha particle, the total momentum still must be zero, meaning momentum of the alpha particle equals the negative of the momentum of the remaining nucleus.
02

Calculate the momentum of the alpha particle

The momentum of the alpha particle can be calculated using its kinetic energy. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Rearranging for velocity, \( v = \sqrt{\frac{2KE}{m}} \). The momentum \( p = mv \). Substitute \( KE = 1.23 \times 10^{-12} \mathrm{J} \) and \( m = 6.65 \times 10^{-27} \mathrm{kg} \) to find \( p \).
03

Calculate the recoil velocity of the nucleus

Use momentum conservation: \( p_{\text{nucleus}} = -p_{\text{alpha}} \). Since \( p = mv \), and letting \( M \) be the mass of the recoiling nucleus \( ^{210}\text{Pb} \), the velocity of the nucleus is \( v = \frac{p_{\text{alpha}}}{M} \). Substitue the mass of \( ^{210}\text{Pb} \approx 3.48 \times 10^{-25} \mathrm{kg} \) along with \( p_{\text{alpha}} \) obtained.

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Key Concepts

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

Momentum Conservation
In the realm of physics, momentum conservation is a crucial principle, especially during interactions like collisions or decays. Momentum is a vector quantity, which means it has both magnitude and direction. When a system is closed (no external forces acting on it), the total momentum before an event must equal the total momentum after it.

In radioactive decay, before the decay process begins, if the parent nucleus is stationary, the total momentum is initially zero. After an alpha particle is emitted, the system should still have a net momentum of zero. This is achieved as the momentum of the emitted alpha particle is exactly balanced by the momentum of the remaining nucleus, but in the opposite direction.
  • The positive momentum of the alpha particle is cancelled out by the negative momentum of the nucleus.
  • Mathematically, this is expressed as: \( p_{\text{alpha}} + p_{\text{nucleus}} = 0 \).
  • As a result, \( p_{\text{nucleus}} = -p_{\text{alpha}} \).
This provides a solid foundation to solve problems involving radioactive decay, like finding the recoil velocity of a nucleus.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In the context of radioactive decay, when an alpha particle is emitted from a nucleus, it is this kinetic energy that is primarily measured. This is because the energy was originally stored in the potential energy of the bonds within the nucleus and is converted into the kinetic energy of the moving alpha particle.
  • If we know the kinetic energy and mass of an alpha particle, we can determine its velocity using the rearranged formula: \( v = \sqrt{\frac{2KE}{m}} \).
  • This velocity helps calculate the momentum of the particle, an essential step in applying momentum conservation.
Understanding kinetic energy is essential in determining both the velocity and momentum of particles involved.
Recoil Velocity
The recoil velocity is the speed at which a nucleus moves backward after emitting an alpha particle. This concept is analogous to how a gun recoils in the opposite direction when a bullet is fired.

To find the recoil velocity of a nucleus, we rely on the principle of momentum conservation. Once the alpha particle's velocity (and, hence, its momentum) is calculated, the recoil velocity of the nucleus can be determined by balancing the momentum equation.
  • Given the formula \( p = mv \), the recoil velocity \( v_{\text{nucleus}} \) is computed as \( v_{\text{nucleus}} = \frac{p_{\text{alpha}}}{M} \).
  • Here, \( M \) is the mass of the residual nucleus, and \( p_{\text{alpha}} \) is the previously calculated momentum of the alpha particle.
  • This step ensures that the momentum conservation principle holds true.
Analyzing recoil velocity is important to confirm energy transfer and equilibrium in nuclear decay.
Alpha Particle Emission
Alpha particle emission is a type of radioactive decay where an alpha particle, consisting of 2 protons and 2 neutrons, is emitted from the nucleus of an atom. This process results in the formation of a new element with a reduced atomic mass and number.

During this emission, several key changes occur:
  • The parent nucleus loses two protons and two neutrons, reducing its atomic number by 2 and its mass number by 4.
  • The emitted alpha particle carries away energy, primarily in the form of kinetic energy.
  • This emission affects both the mass and energy balance within the nucleus, leading to recoil and the necessity of understanding conservation principles.
By studying alpha particle emission, scientists gain insights into nuclear stability, energy distribution, and the behavior of matter at a subatomic level.

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Most popular questions from this chapter

A \(20.00-\mathrm{kg}\) lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a \(5.00-\mathrm{kg}\) steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

In a volcanic eruption, a 2400 -kg boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands 274 \(\mathrm{m}\) directly north of the point of the explosion. Where will the other fragment land? Neglect any air resistance.

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) , while the other. of mass 65.0 \(\mathrm{kg}\) , is moving to the left at 2.50 \(\mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the velocity of these skaters just after they collide?

In beta decay, a nucleus emits an electron. A 210 Bi (bismuth) nucleus at rest undergoes beta decay to \(^{210} \mathrm{Po}\) (polonium). Suppose the emitted electron moves to the right with a momentum of \(5.60 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) . The 210 \(\mathrm{Po}\) nucleus, with mass \(3.50 \times 10^{-25} \mathrm{kg},\) recoils to the left at a speed of \(1.14 \times 10^{-3} \mathrm{m} / \mathrm{s}\) Momentum conservation requires that a second particle, called an antineutrino, must also be emitted. Calculate the magnitude and direction of the momentum of the antineutrino that is emitted in this decay.

\(\mathrm{A}^{232} \mathrm{Th}\) (thorium) nucleus at rest decays to a \(^{228} \mathrm{Ra}\) (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is \(6.54 \times 10^{-13} \mathrm{J}\) . An alpha particle has 1.76\(\%\) of the mass of a 28 Ra nucleus. Calculate the kinetic energy of (a) the recoiling 28 Ra nucleus and \((b)\) the alpha particle.

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