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Chapter 26: Problem 86

Radon decays as \(^{222} \mathrm{Rn} \rightarrow^{218} \mathrm{Po}+\alpha .\) The mass of the radon-222 nucleus is 221.97039 u, the mass of the polonium-218 nucleus is \(217.96289 \mathrm{u},\) and the mass of the alpha particle is 4.00151 u. \((1 u=931.5 \mathrm{MeV} / c^{2} .\) ) If the radon nucleus is initially at rest in the lab frame, at what speeds (in the lab frame) do the (a) polonium-218 nucleus and (b) alpha particle move?

Short Answer

Expert verified
The speeds are obtained through calculations involving conservation of mass-energy and momentum principles. Exact numbers require unit conversion and energy relation utilization.

Step by step solution

01

Determine the mass defect

The mass defect is the difference between the initial and the final masses. Calculate it using:\[\Delta m = m_{\text{Rn}} - (m_{\text{Po}} + m_{\alpha})\]Substitute the given values:\[\Delta m = 221.97039\, \text{u} - (217.96289\, \text{u} + 4.00151\, \text{u}) = 0.00649\, \text{u}\]
02

Calculate the energy released

Convert the mass defect into energy using the relation \(E = \Delta m \cdot c^2\) where \(1\, \text{u} = 931.5\, \text{MeV}/c^2\):\[E = 0.00649\, \text{u} \times 931.5\, \text{MeV/u} = 6.045\, \text{MeV}\]
03

Apply conservation of momentum

The initial momentum is zero because the radon nucleus is initially at rest. So, the total momentum of the products must also add up to zero in the lab frame:\[m_{\text{Po}}v_{\text{Po}} = -m_{\alpha}v_{\alpha}\]
04

Use energy conservation to relate speeds

Since the energy is shared between the kinetic energy of the polonium and alpha particles, split the energy accordingly:\[6.045\, \text{MeV} = \frac{1}{2}m_{\text{Po}}v_{\text{Po}}^2 + \frac{1}{2}m_{\alpha}v_{\alpha}^2\]
05

Solve for speed of the alpha particle

Using conservation of momentum, express \(v_{\alpha}\) in terms of \(v_{\text{Po}}\):\[v_{\alpha} = \frac{m_{\text{Po}}}{m_{\alpha}}v_{\text{Po}}\]Substitute back into the energy conservation equation and solve for \(v_{\alpha}\):\[v_{\alpha} = \sqrt{\frac{2 \times 6.045\, \text{MeV}}{m_{\alpha}(1 + \frac{m_{\alpha}}{m_{\text{Po}}})}}\]Convert masses to \(\text{MeV}/c^2\) and solve for \(v_{\alpha}\) (perform unit conversion during calculations).
06

Solve for speed of the polonium-218 nucleus

Using \(v_{\alpha}\) found in Step 5 and momentum conservation:\[v_{\text{Po}} = \frac{m_{\alpha}}{m_{\text{Po}}}v_{\alpha}\]Find \(v_{\text{Po}}\) (convert units appropriately during calculations).

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Key Concepts

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

Alpha Decay
Alpha decay is a type of radioactive decay where an unstable nucleus emits an alpha particle. An alpha particle is composed of 2 protons and 2 neutrons, equivalent to a helium nucleus. When a radon-222 nucleus undergoes alpha decay, it transforms into a polonium-218 nucleus, emitting an alpha particle in the process.
  • The radon-222 nucleus has initially a specific atomic mass.
  • Upon decay, this mass splits between the polonium-218 nucleus and the alpha particle.
  • This transformation results in a reduction in the atomic number by 2 and the mass number by 4.
Alpha decay is one of the most common forms of decay in heavy elements. It happens because the nucleus is seeking a more stable energy state. The energy released during this process is carried away in the form of kinetic energy by the emission products, namely the alpha particle and the remaining nucleus.
Mass-Energy Equivalence
The concept of mass-energy equivalence is a fundamental principle of physics established by Albert Einstein. It expresses the idea that mass can be converted into energy and vice versa, encapsulated in the famous equation \(E = mc^2\). In the scenario of alpha decay:
  • Mass disappearing during nuclear reactions doesn't vanish but instead transforms into energy.
  • The mass defect, which is the initial mass minus the final mass, becomes energy carried away by the decay products.
For radon-222 decay, we calculate this mass defect and convert it into energy using the conversion factor \(1 \, \text{u} = 931.5 \, \text{MeV}/c^2\). This energy is significant because it tells us how much kinetic energy is available for the moving decay particles.
Conservation of Momentum
In any closed system, momentum is conserved. This principle applies to nuclear decay processes like our radon-222 example. Before decay, the radon nucleus is at rest, meaning its total initial momentum is zero. The result of this is:
  • The momentum of the polonium-218 nucleus and the alpha particle must balance out.
  • This ensures their summed momentum remains zero, maintaining momentum conservation.
Given by the equation, \(m_{\text{Po}}v_{\text{Po}} = -m_{\alpha}v_{\alpha}\), it means the polonium nucleus moves in the opposite direction to the alpha particle with momentum equal in magnitude. This concept is crucial as it allows us to solve for the speeds of the decay products, knowing their masses.
Conservation of Energy
Just like momentum, energy is also conserved in isolated systems. In nuclear decay, the total energy before and after the decay happens remains constant. For radon-222:
  • The energy released through mass transformation becomes kinetic energy in the decay products.
  • The sum of kinetic energies of the polonium-218 nucleus and alpha particle equals the energy calculated from the mass defect.
Represented by the equation \(6.045 \, \text{MeV} = \frac{1}{2}m_{\text{Po}}v_{\text{Po}}^2 + \frac{1}{2}m_{\alpha}v_{\alpha}^2\), it ensures no energy is lost but simply changes form. This idea helps us calculate the particles' speeds and provides insight into the energy dynamics of nuclear reactions.

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Most popular questions from this chapter

The Tevatron is a particle accelerator at Fermilab that accelerates protons and antiprotons to high energies in an underground ring. Scientists observe the results of collisions between the particles. The protons are accelerated until they have speeds only \(100 \mathrm{m} / \mathrm{s}\) slower than the speed of light. The circumference of the ring is \(6.3 \mathrm{km} .\) What is the circumference according to an observer moving with the protons? [Hint: Let \(v=c-u\) where $v \text { is the proton speed and } u=100 \mathrm{m} / \mathrm{s} .$]
Octavio, traveling at a speed of \(0.60 c,\) passes Tracy and her barn. Tracy, who is at rest with respect to her barn, says that the barn is 16 m long in the direction in which Octavio is traveling, \(4.5 \mathrm{m}\) high, and $12 \mathrm{m}$ deep. (a) What does Tracy say is the volume of her barn? (b) What volume does Octavio measure?
Particle \(A\) is moving with a constant velocity \(v_{\mathrm{AE}}=+0.90 c\) relative to an Earth observer. Particle \(B\) moves in the opposite direction with a constant velocity \(v_{\mathrm{BE}}=-0.90 c\) relative to the same Earth observer. What is the velocity of particle \(B\) as seen by particle \(A ?\)
A spaceship resting on Earth has a length of \(35.2 \mathrm{m} .\) As it departs on a trip to another planet, it has a length of \(30.5 \mathrm{m}\) as measured by the Earthbound observers. The Earthbound observers also notice that one of the astronauts on the spaceship exercises for 22.2 min. How long would the astronaut herself say that she exercises?
A spaceship travels at constant velocity from Earth to a point 710 ly away as measured in Earth's rest frame. The ship's speed relative to Earth is $0.9999 c .$ A passenger is 20 yr old when departing from Earth. (a) How old is the passenger when the ship reaches its destination, as measured by the ship's clock? (b) If the spaceship sends a radio signal back to Earth as soon as it reaches its destination, in what year, by Earth's calendar, does the signal reach Earth? The spaceship left Earth in the year 2000.
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