Question

A nucleus of \({ }^{210}{ }_{84}\) Po originally at rest emits \(\alpha\) -particle with speed \(\mathrm{v}\) what will be the recoil speed of the daughter nucleus (A) \([\mathrm{v} /(214)]\) (B) \([(4 \mathrm{v}) /(214)]\) (C) \([(4 \mathrm{v}) /(206)]\) (D) \([\mathrm{v} /(206)]\)

Short answer

The recoil speed of the daughter nucleus is (C) \([(4 \mathrm{v}) /(206)]\).

Step-by-step solution

Identify the given information

Given information: 1. The initial nucleus is Po-210: It has 210 nucleons and 84 protons (mass number A = 210; atomic number Z = 84) 2. An α-particle is emitted: An α-particle consists of 2 protons and 2 neutrons (mass number A = 4; atomic number Z = 2) 3. The speed of the emitted α-particle is v

Calculate the mass of the daughter nucleus

Since the α-particle has a mass number of 4, the mass number of the daughter nucleus is the difference between the parent nucleus and the α-particle: A_daughter = A_parent - A_alpha = 210 - 4 = 206 The daughter nucleus has 206 nucleons and 82 protons (mass number A = 206; atomic number Z = 82)

Apply the conservation of momentum principle

Before the α-particle emission, the system is at rest, and its total linear momentum is zero. After the emission, the sum of the linear momentum of the α-particle and daughter nucleus should still be zero. Let v_daughter be the recoil speed of the daughter nucleus. Then, applying the conservation of momentum principle: \(mass_{daughter} \cdot v_{daughter} = mass_{alpha} \cdot v_{alpha}\) Since the masses of the α-particle and daughter nucleus are proportional to their mass numbers, we can rewrite the equation as: \(A_{daughter} \cdot v_{daughter} = A_{alpha} \cdot v_{alpha}\)

Solve for the recoil speed of the daughter nucleus

Now we can solve for v_daughter: \(v_{daughter} = (A_{alpha} \cdot v_{alpha}) / A_{daughter}\) Substitute the values we found earlier: \(v_{daughter} = (4 \cdot v) / 206\) Comparing this to the given answer choices, we find that the correct answer is: (C) \([(4 \mathrm{v}) /(206)]\)

Key concepts

Momentum Conservation

In physics, the principle of momentum conservation is fundamental when analyzing collisions and decays, including nuclear decays. In a closed system with no external forces, the total momentum before an event is equal to the total momentum after the event. This principle helps to determine unknown velocities after interactions, such as the emission of particles during nuclear decay.

For example, in the decay of a nucleus where an alpha particle is emitted, the system initially is at rest, meaning its total momentum is zero. When the alpha particle is emitted, it gains momentum. To conserve the total momentum of the system, the remaining daughter nucleus must also have an equal and opposite momentum. This ensures momentum conservation overall, balancing the initial state where no motion or external forces were present.

Understanding this principle helps predict the resulting speed and direction of nuclei or particles involved in nuclear reactions and can be calculated by equating the momentum of the emitted particle to the momentum of the recoiling nucleus.

Alpha Particle Emission

Alpha particle emission is a type of nuclear decay where an unstable nucleus releases an alpha particle to become more stable. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons, and carries a positive charge.

When a nucleus emits an alpha particle:

• It reduces its atomic mass by 4 units (since the alpha particle itself has a mass number of 4).
• It reduces its atomic number by 2, changing the element to a different one in the periodic table.
• The original nucleus becomes the 'daughter nucleus' post-decay.

Alpha particle emissions are common in heavy elements, allowing them to release energy and transition to a more stable form. The energy and speed of the emitted alpha particles are typically high, making them significant in radiation protection and nuclear chemistry studies.

Daughter Nucleus

When a nucleus undergoes a process like alpha decay, the original nucleus transforms into a different nucleus known as the daughter nucleus.

• The mass number of the daughter nucleus is reduced from the original by 4, reflecting the mass of the alpha particle emitted.
• The atomic number decreases by 2 due to the loss of two protons as part of the alpha particle.
• In addition to the change in atomic composition, the daughter nucleus often gains additional stability and sits as a lower-energy state compared to its parent.

In the given example of polonium ( ^{210}_{84} ext{{Po}}) decaying by emitting an alpha particle, the daughter nucleus formed is lead ( ^{206}_{82} ext{{Pb}}). This transformation is critical in understanding how elements naturally transmute over time and the processes involved in nuclear reactions.

Recoil Speed Calculation

Calculating the recoil speed of a daughter nucleus is a vital step when analyzing nuclear decays, using the principle of conservation of momentum. Consider the scenario where a nucleus emits an alpha particle.

• Initially, the nucleus is at rest, so total momentum before the emission is zero.
• Upon emission, the alpha particle gains speed and momentum in one direction, requiring the daughter nucleus to move in the opposite direction to conserve total momentum.

To find the recoil speed of the daughter nucleus: \[v_{\text{daughter}} = \left(\frac{A_{\text{alpha}} \cdot v_{\alpha}}{A_{\text{daughter}}}\right)\]Where:

• \(A_{\text{alpha}}\) is the mass number of the alpha particle.
• \(v_{\alpha}\) is the speed of the alpha particle.
• \(A_{\text{daughter}}\) is the mass number of the daughter nucleus after emission.

For our example, inserting values for the alpha particle and daughter nucleus, the recoil speed can be calculated as:\[v_{\text{daughter}} = \frac{4 \cdot v}{206}\]This formula demonstrates how the change in mass affects the recoil speed of the daughter nucleus, which is usually much smaller in comparison to the emitted alpha particle's speed.