
A nucleus of mass $m + \Delta m$ is at rest and decays into two daughter nuclei of equal mass each
$\dfrac{M}{2}$. The speed of light is $c$. The speed of daughter nuclei is
a. $c\dfrac{{\Delta m}}{{M + \Delta m}}$
b. $c\sqrt {\dfrac{{2\Delta m}}{M}} $
C. $c\sqrt {\dfrac{{\Delta m}}{M}} $
d. $c\sqrt {\dfrac{{\Delta m}}{{M + \Delta m}}} $
Answer
146.7k+ views
Hint: In this question, first use the law of conservation of momentum and then find the velocity of the daughter nuclei is the same. Then find the total kinetic energy of the two daughter nuclei and mass defect and then find the velocity of the two daughter nuclei.
Complete step by step answer:
A nucleus decays into two parts and the mass of each nucleus is $\dfrac{M}{2}$. The mass of the parent nucleus is $M + \Delta m$ .
Let us assume that the speed of daughter nuclei is ${V_1}$ and ${V_2}$ respectively.
Hence conservation of momentum, $\dfrac{M}{2}{V_1} = \dfrac{M}{2}{V_2} \Rightarrow {V_1} = {V_2}$
Now the mass defect is $M + \Delta m - \left( {\dfrac{M}{2} + \dfrac{M}{2}} \right) = \Delta m$
As the product of mass defect and the square of the speed of light is the total kinetic energy.
So, the kinetic energy of the two daughter nuclei is ${E_1} = \dfrac{1}{2}.\dfrac{M}{2}.{V_1}^2$ and ${E_2} = \dfrac{1}{2}.\dfrac{M}{2}.{V_2}^2$
Hence the total kinetic energy of the two daughter nuclei is ${E_1} + {E_2} = \dfrac{1}{2}.\dfrac{M}{2}.{V_1}^2 + \dfrac{1}{2}.\dfrac{M}{2}.{V_2}^2$
As the velocity of the two daughter nuclei is same ${V_1} = {V_2}$
Thus, the total kinetic energy of the two daughter nuclei is $\dfrac{M}{2}{V_1}^2$ .
$ \Rightarrow \Delta m{c^2} = \dfrac{M}{2}{V_1}^2$
$\therefore {V_1} = c\sqrt {\dfrac{{2\Delta m}}{M}} $
The speed of the two daughter nuclei is $c\sqrt {\dfrac{{2\Delta m}}{M}} $.
Hence option (b) is the correct answer.
Note: As we know that the law of conservation of momentum states that in an isolated system, when the two objects collide with each other the total momentum of two objects before the collision is equal to the total momentum after the collision. Momentum is neither destroyed nor created; it transforms into one form to another.
Complete step by step answer:
A nucleus decays into two parts and the mass of each nucleus is $\dfrac{M}{2}$. The mass of the parent nucleus is $M + \Delta m$ .
Let us assume that the speed of daughter nuclei is ${V_1}$ and ${V_2}$ respectively.
Hence conservation of momentum, $\dfrac{M}{2}{V_1} = \dfrac{M}{2}{V_2} \Rightarrow {V_1} = {V_2}$
Now the mass defect is $M + \Delta m - \left( {\dfrac{M}{2} + \dfrac{M}{2}} \right) = \Delta m$
As the product of mass defect and the square of the speed of light is the total kinetic energy.
So, the kinetic energy of the two daughter nuclei is ${E_1} = \dfrac{1}{2}.\dfrac{M}{2}.{V_1}^2$ and ${E_2} = \dfrac{1}{2}.\dfrac{M}{2}.{V_2}^2$
Hence the total kinetic energy of the two daughter nuclei is ${E_1} + {E_2} = \dfrac{1}{2}.\dfrac{M}{2}.{V_1}^2 + \dfrac{1}{2}.\dfrac{M}{2}.{V_2}^2$
As the velocity of the two daughter nuclei is same ${V_1} = {V_2}$
Thus, the total kinetic energy of the two daughter nuclei is $\dfrac{M}{2}{V_1}^2$ .
$ \Rightarrow \Delta m{c^2} = \dfrac{M}{2}{V_1}^2$
$\therefore {V_1} = c\sqrt {\dfrac{{2\Delta m}}{M}} $
The speed of the two daughter nuclei is $c\sqrt {\dfrac{{2\Delta m}}{M}} $.
Hence option (b) is the correct answer.
Note: As we know that the law of conservation of momentum states that in an isolated system, when the two objects collide with each other the total momentum of two objects before the collision is equal to the total momentum after the collision. Momentum is neither destroyed nor created; it transforms into one form to another.
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
