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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.

Short Answer

Expert verified
The antineutrino's momentum is approximately \(-5.60 \times 10^{-22} \text{ kg} \cdot \text{ m/s}\), directed to the left.

Step by step solution

01

Understand the concepts

When an object is at rest, its momentum is zero. In a closed system, the total initial momentum equals the total final momentum. Here, a 210 Bi nucleus decays, and since it is initially at rest, it has zero initial momentum. After the decay, we need to calculate the sum of the momentum of the electron, the newly formed 210 Po nucleus, and the antineutrino to ensure momentum conservation.
02

Calculate the final momentum of Polonium

The momentum of the polonium nucleus can be calculated using the formula: \[ p_{Po} = m_{Po} \cdot v_{Po} \] Given that the mass of polonium, \( m_{Po} = 3.50 \times 10^{-25} \text{kg} \), and its speed \( v_{Po} = 1.14 \times 10^{-3} \text{m/s} \), \[ p_{Po} = 3.50 \times 10^{-25} \times 1.14 \times 10^{-3} = 3.99 \times 10^{-28} \text{kg} \cdot \text{m/s} \] This momentum is directed to the left.
03

Apply conservation of momentum

Since the initial momentum of the system is zero, the sum of the final momenta of all particles should also be zero. The contributions to the momentum are:1. Electron momentum \( p_{e} = 5.60 \times 10^{-22} \text{ kg} \cdot \text{ m/s} \) (right)2. Polonium momentum \( p_{Po} = 3.99 \times 10^{-28} \text{ kg} \cdot \text{ m/s} \) (left)3. Antineutrino momentum \( p_{\overline{u}} \) (unknown)The conservation equation is:\[ 0 = 5.60 \times 10^{-22} - 3.99 \times 10^{-28} + p_{\overline{u}} \]
04

Solve for antineutrino momentum

Rearranging the conservation equation to solve for the antineutrino momentum:\[ p_{\overline{u}} = -5.60 \times 10^{-22} + 3.99 \times 10^{-28} \]Given the very small magnitude of the polonium momentum relative to the electron's, the antineutrino's momentum will be approximately:\[ p_{\overline{u}} \approx -5.60 \times 10^{-22} \text{ kg} \cdot \text{ m/s} \]The negative sign indicates that the antineutrino's momentum is directed opposite to the electron's direction, i.e., to the left.

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a fundamental principle in physics that says that the total momentum of a closed system is constant, provided no external forces act on it. This means that if you have an object, or a group of objects, their total momentum before any event must be equal to their total momentum after the event. In beta decay, a nucleus emitting particles maintains its overall momentum. When a 210 Bi nucleus at rest decays, the initial momentum is zero. After the decay, the combined momentum of the emitted electron, the recoiling polonium nucleus, and the antineutrino must sum up to zero to satisfy the conservation of momentum. This is why even if a particle moves in one direction, another must move oppositely. This keeps the system balanced and its total momentum unchanged.
Antineutrino
During beta decay, an antineutrino is a crucial emitted particle. Neutrinos and antineutrinos are elementary particles with very small masses that rarely interact with other matter. In the context of beta decay, the antineutrino is essential because it helps conserve both momentum and energy. For our exercise, the antineutrino has a momentum opposite to that of the emitted electron. This ensures that overall, the particles behave according to physical laws. Antineutrinos, like neutrinos, are neutral (no electric charge) and almost invisible to the surrounding environment. They travel through space with high energy, taking away part of the momentum from the beta decay process. Despite their elusive nature, they are fundamental in maintaining the conservation laws in nuclear reactions such as this.
Electron Emission
Electron emission is an integral part of beta decay. It occurs when a neutron in a nucleus transforms into a proton. This transformation process emits an electron and an antineutrino. The electron, often referred to as a beta particle in this context, leaves the nucleus carrying away energy and momentum. In this exercise, the momentum of the emitted electron is known, and it's significant enough to necessitate a balance by another particle moving in the opposite direction. The high-speed electron is the reason we must look at other particles, like the antineutrino and the recoiling nucleus, to understand where all the initial momentum went. This emission is one of the key processes that bring about changes within the nucleus, effectively altering the element by increasing its atomic number by one.
Nuclear Physics
Nuclear physics is the field that studies the components and behavior of the nucleus. It explores phenomena like beta decay, which plays a crucial role in changing the composition of an atom on a sub-atomic level. Within this realm, scientists are particularly interested in understanding how particles interact to form nuclei, the forces that hold nuclei together, and the processes by which unstable nuclei transform into stable configurations. In beta decay, such as the one described in the exercise, a neutron inside an atom's nucleus decays into a proton. This leads to the emission of both an antineutrino and an electron. By studying these processes, nuclear physics can explain the stability of matter and the nature of radioactive substances. This area provides insight into how elements form, transform, and eventually stabilize over time, contributing to our broader understanding of atomic interactions and fundamental forces.

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Most popular questions from this chapter

At one instant, the center of mass of a system of two particles is located on the \(x\) -axis at \(x=2.0 \mathrm{m}\) and has a velocity of \((5.0 \mathrm{m} / \mathrm{s})\) ) One of the particles is at the origin. The other particle has a mass of 0.10 \(\mathrm{kg}\) and is at rest on the \(x\) -axis at \(x=8.0 \mathrm{m}\) . (a) What is the mass of the particle at the origin? (b) Calculate the total momentum of this system. (c) What is the velocity of the particle at the origin?

A \(1050-\mathrm{kg}\) sports car is moving westbound at 15.0 \(\mathrm{m} / \mathrm{s}\) on a level road when it collides with a 6320 \(\mathrm{kg}\) truck driving east on the same road at 10.0 \(\mathrm{m} / \mathrm{s}\) . The two vehicles remain locked together after the collision. (a) What is the velocity (magnitude and direction) of the two vehicles just after the collision? (b) At what speed should the truck have been moving so that it and car are both stopped in the collision? (c) Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an clastic collision with the floor and then rebounds to make an elastic collision with the still- descending small ball. Just before the collision between the two balls, the large ball is moving upward with velocity \(\overrightarrow{\boldsymbol{v}}\) and the small ball has velocity \(-\overrightarrow{\boldsymbol{v}}\) . (Do you see why? Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large hall? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?

A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in 50.0 \(\mathrm{s}\) and the relative speed of the exhaust gas is \(v_{\mathrm{ex}}=2100 \mathrm{m} / \mathrm{s}\) , what must the mass ratio \(m_{0} / m\) be for a foral speed \(v\) of 8.00 \(\mathrm{km} / \mathrm{s}\) (about equal to the orbital speed of an earth satellite)?

A 12.0 kg shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of 150 \(\mathrm{m} / \mathrm{s}\) . When it is at its highest point, the shell exploded into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Assume that air resistance can be ignored. If the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land how much energy was released in the explosion?

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