Answer
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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.
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