Question

A neutron makes a head-on elastic collision with a lead nucleus. The ratio of nuclear mass to neutron mass is 206. The fractional change in kinetic energy of a neutron is:

• A. 3% increase
• B. 2% decrease
• C. 2% increase
• D. 3% decrease

Hint:

In elastic collisions, the fractional change in kinetic energy can be calculated using the formula \( \frac{(m_1 - m_2)^2}{(m_1 + m_2)^2} \), where \( m_1 \) and \( m_2 \) are the masses of the two colliding objects.

Answer 1

The Correct Option is B

This problem involves an elastic collision between a neutron and a lead nucleus. In elastic collisions, both momentum and kinetic energy are conserve(D) The key to solving this problem is understanding how the change in kinetic energy depends on the masses involve(D) In a head-on elastic collision between two bodies, the fractional change in kinetic energy for the body of mass \( m_1 \) (the neutron) when it collides with a body of mass \( m_2 \) (the lead nucleus) is given by the formula: \[ \Delta KE = \frac{(m_1 - m_2)^2}{(m_1 + m_2)^2} \] Where: - \( m_1 \) is the mass of the neutron, - \( m_2 \) is the mass of the lead nucleus, - The ratio of nuclear mass to neutron mass is given as \( \frac{m_2}{m_1} = 206 \), so \( m_2 = 206 m_1 \). Substituting \( m_2 = 206 m_1 \) into the formula: \[ \Delta KE = \frac{(m_1 - 206 m_1)^2}{(m_1 + 206 m_1)^2} = \frac{(1 - 206)^2}{(1 + 206)^2} \] \[ \Delta KE = \frac{(-205)^2}{(207)^2} = \frac{42025}{42849} \approx 0.982 \] Thus, the fractional change in kinetic energy is approximately 0.982, which represents a decrease in the kinetic energy of the neutron. This is a decrease of about 2%. Thus, the correct answer is Option (B): "2% decrease".