An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of $8 : 27$. The ratio of the radii of the nuclei (assumed to be spherical) is :
To solve this problem, let us consider the principles of conservation of momentum and the properties of spherical nuclei.
The problem states that a heavy nucleus at rest breaks into two smaller nuclei moving with velocities in the ratio 8 : 27.
We know from the conservation of momentum that the momentum before the break-up (which is zero, as the nucleus is at rest) must equal the total momentum after the break-up.
Let the masses of the two smaller nuclei be m_1 and m_2, and their velocities be v_1 and v_2 respectively.
According to conservation of momentum:
m_1 v_1 = m_2 v_2
Given \frac{v_1}{v_2} = \frac{8}{27}, we can express this relation as:
v_1 = \frac{8}{27}v_2
Substituting into the momentum equation:
m_1 \frac{8}{27}v_2 = m_2 v_2
Simplifying yields:
m_1 = \frac{27}{8}m_2
Thus, the mass ratio m_1 : m_2 = 27 : 8.
The volume of a spherical nucleus is given by:
V = \frac{4}{3}\pi r^3
where r is the radius.
Assuming the nuclei are spherical and made of the same material, we can express mass in terms of volume, which gives:
\frac{m_1}{m_2} = \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3
Substituting the mass ratio:
\frac{27}{8} = \left(\frac{r_1}{r_2}\right)^3
Taking the cube root on both sides:
\frac{r_1}{r_2} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}
Hence, the ratio of the radii of the two nuclei is 3 : 2.
