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) \([(4 \mathrm{v}) /(206)]\)