Physics · JEE | WORK, ENERGY AND POWER
हिन्दीWhat is the fractional kinetic energy lost \( f_1 = \frac{K_{1f}}{K_{1i}} \) by a neutron of mass \( m_1 \) in an elastic collision with a moderating nucleus of mass \( m_2 \)?
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Correct answer · Option D
\( f_1 = (\frac{m_1 - m_2}{m_1 + m_2})^2 \)
Explanation
Using \( K = \frac{1}{2}mv^2 \) and \( v_{1f} = (\frac{m_1 - m_2}{m_1 + m_2})v_{1i} \), the fractional kinetic energy retained is \( f_1 = (\frac{m_1 - m_2}{m_1 + m_2})^2 \).