Question

A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer 1

Given

A nucleus is initially at rest in the laboratory frame. Let its mass be \( M \). It disintegrates into two smaller nuclei of masses \( m_1 \) and \( m_2 \), moving with velocities \( \vec{v}_1 \) and \( \vec{v}_2 \), respectively.

Conservation of Linear Momentum

Since the nucleus is at rest initially, its initial linear momentum is zero:

\( \vec{p}_{\text{initial}} = 0 \)

After disintegration, the total linear momentum of the two product nuclei is

\( \vec{p}_{\text{final}} = m_1 \vec{v}_1 + m_2 \vec{v}_2 \)

In absence of external forces, linear momentum of the system is conserved:

\( \vec{p}_{\text{initial}} = \vec{p}_{\text{final}} \Rightarrow 0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 \)

Direction of Motion of the Products

Rearranging the momentum-conservation equation:

\( m_1 \vec{v}_1 = -\, m_2 \vec{v}_2 \)

or

\( \vec{v}_2 = -\, \dfrac{m_1}{m_2}\, \vec{v}_1 \)

The minus sign shows that the velocity vectors \( \vec{v}_1 \) and \( \vec{v}_2 \) are in opposite directions. Therefore, the two smaller nuclei must move in opposite directions after the disintegration.

Physical Explanation

• No external force acts on the nucleus–products system, so its total momentum must remain zero.
• With only two product nuclei, the only way to keep the vector sum of their momenta zero is for them to have equal and opposite momenta, i.e., they move along the same line in opposite directions.