JEE Physics 2023 Question with Solution

Given: Ratio of nuclear sizes is $r_1 : r_2 = 1 : 2^{1/3}$. Ratio of speeds is $V_1 : V_2 = n : 1$.

Find: The value of $n$.

Assuming the original nucleus is initially at rest, conservation of linear momentum gives

$$m_1 V_1 = m_2 V_2$$

So,

$$\frac{V_1}{V_2} = \frac{m_2}{m_1}$$

For nuclear fragments of the same density, mass is proportional to volume, hence

$$\frac{m_2}{m_1} = \frac{r_2^3}{r_1^3}$$

Given

$$\frac{r_2}{r_1} = 2^{1/3}$$

Therefore,

$$\frac{m_2}{m_1} = \left(\frac{r_2}{r_1}\right)^3 = \left(2^{1/3}\right)^3 = 2$$

Hence,

$$\frac{V_1}{V_2} = 2$$

So $V_1 : V_2 = 2 : 1$, therefore the value of $n$ is $2$.

Given: $r_1 : r_2 = 1 : 2^{1/3}$ and $V_1 : V_2 = n : 1$.

Find: The value of $n$.

The two fragments move in opposite directions after disintegration. Since the initial nucleus is at rest, total momentum after disintegration must remain zero. Hence their momenta are equal in magnitude:

$$m_1 V_1 = m_2 V_2$$

This gives

$$\frac{V_1}{V_2} = \frac{m_2}{m_1}$$

Now use the fact that nuclear density is approximately constant. Therefore mass is proportional to volume, and volume is proportional to cube of radius:

$$m \propto r^3$$

Thus,

$$\frac{m_2}{m_1} = \frac{r_2^3}{r_1^3}$$

From the given size ratio,

$$\frac{r_2}{r_1} = 2^{1/3}$$

Cubing both sides,

$$\frac{r_2^3}{r_1^3} = 2$$

So,

$$\frac{m_2}{m_1} = 2$$

Therefore,

$$\frac{V_1}{V_2} = 2$$

Comparing with $n : 1$, we get $n = 2$.

Therefore, the required numerical value is $2$.