A nucleus of mass M at rest splits into two parts having masses $\frac{M′}{3}$ and $\frac{2M′}{3}(M′<M).$

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

To solve this problem, we have to understand the concept of conservation of momentum. When a nucleus at rest splits into two parts, the sum of their momenta must be zero because initially, the total momentum is zero.

Let's consider the two parts after the split:

Mass of the first part: $\frac{M'}{3}$
Mass of the second part: $\frac{2M'}{3}$

Let the velocities of these two parts be v_1 and v_2 respectively.

According to the conservation of linear momentum:

$\frac{M'}{3} \cdot v_1 + \frac{2M'}{3} \cdot v_2 = 0$

We can rearrange this equation to find the relationship between the velocities:

v_1 = -2v_2

This negative sign indicates that the velocities are in opposite directions.

To find the ratio of the kinetic energies, we use the kinetic energy formula:

KE = \frac{1}{2} mv^2

Let's calculate the kinetic energies for both parts:

Kinetic energy of the first part: KE_1 = \frac{1}{2} \cdot \frac{M'}{3} \cdot (v_1)^2 = \frac{1}{2} \cdot \frac{M'}{3} \cdot (4v_2^2)
Kinetic energy of the second part: KE_2 = \frac{1}{2} \cdot \frac{2M'}{3} \cdot (v_2)^2

Taking the ratio of these kinetic energies:

$\frac{KE_1}{KE_2} = \frac{\frac{1}{2} \cdot \frac{M'}{3} \cdot 4v_2^2}{\frac{1}{2} \cdot \frac{2M'}{3} \cdot v_2^2} = \frac{\frac{4M'}{3}}{\frac{2M'}{3}} = \frac{4}{2} = 2$

However, remember that v_1 = -2v_2, so substituting this correctly:

$\frac{KE_1}{KE_2} = \frac{\frac{4M'}{3}}{\frac{2M'}{3}} \cdot (\frac{v_2^2}{v_1^2}) = \frac{4}{2} \cdot (\frac{v_2^2}{4v_2^2}) = 1$

Therefore, the kinetic energies of the two parts are in the ratio 1:1.

The correct answer is 1:1.

Answer 2