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Chapter 18: Problem 2471

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

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

Step by step solution

01

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
02

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)
03

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: \(A_{daughter} \cdot v_{daughter} = A_{alpha} \cdot v_{alpha}\)
04

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

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Most popular questions from this chapter

The binding energy Per nucleon of \({ }_{8} \mathrm{O}^{16}\) is $7.97 \mathrm{MeV}\( and that of \)_{8} \mathrm{O}^{17}\( is \)7.75 \mathrm{MeV}$ The energy \((\mathrm{in}-\mathrm{MeV})\) required to remove a neutron from $_{8} \mathrm{O}^{17}$ is (A) \(3.65\) (B) \(7.86\) (C) \(3.52\) (D) \(4.23\)

The innermost orbit of the hydrogen atom has a radius \(0.53 \mathrm{~A}\). what is radius of \(2^{\text {nd }}\) orbit is ? (A) \(2.12 \AA\) (B) \(1.06 \AA\) (C) \(21.2 \AA\) (D) \(10.6 \AA\)

Match column I and II column I \(\frac{\text { column I }}{\text { fnucleus }}\) (a) size of (b) number of Proton in a nucleus column II (p) Z (q) \(10^{-15} \mathrm{~m}\) (r) \((\mathrm{A}-\mathrm{Z})\) (s) \(10^{-10} \mathrm{~m}\) 0 (c) size of Atom (d) Number of neutrons in a nucleus (A) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{r}$ (C) $\mathrm{a} \rightarrow \mathrm{s}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{r}$

The wave length of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number \(\mathrm{Z}\) of hydrogen like ion is (A) 1 (B) 2 (C) 3 (D) 4

If a hydrogen atom emits a Photon of wave length \(\lambda\). the recoil speed of the atom of mass \(\mathrm{m}\) is given by (A) \((\mathrm{h} / \mathrm{m} \lambda)\) (B) \((\mathrm{mh} / \lambda)\) (C) \(\operatorname{mh} \lambda\) (D) \((\mathrm{m} \lambda / \mathrm{h})\)
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