Question 328: Atoms & Nuclei

Question

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $$1: 2^{1 / 3}$$. Their respective speed have a ratio of $$n: 1$$. The value of $n$ is __________.

Accepted Answer

Let the masses of the two nuclear parts be $$m_1$$ and $$m_2$$, and their respective speeds be $$v_1$$ and $$v_2$$. According to the problem, the ratio of their nuclear sizes is $$1 : 2^{1/3}$$. Since the nuclear size is proportional to the cube root of the mass, we can write:
$$
\frac{m_1}{m_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2}
$$
Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:
$$
m_1 v_1 = m_2 v_2
$$
From the problem statement, the ratio of their respective speeds is $$n : 1$$, so we can write:
$$
v_1 = n \cdot v_2
$$
Substitute the expression for $$v_1$$ into the momentum conservation equation:
$$
m_1 (n \cdot v_2) = m_2 v_2
$$
We know the mass ratio, so substitute that into the equation:
$$
\frac{1}{2} m_2 (n \cdot v_2) = m_2 v_2
$$
Divide both sides by $$m_2 v_2$$:
$$
\frac{1}{2} n = 1
$$
Now, solve for $$n$$:
$$
n = 2
$$
Thus, the value of $$n$$ is 2.

Atoms & Nuclei

A nucleus disintegrates into two nuclear parts, in such a way that ratio of their nuclear sizes is $1: 2^{1 / 3}. Their respective speed have a ratio of n: 1. The value of n$ is __________.

Integer / Numeric Answer Type