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(B) In a head-on elastic collision,the fractional energy loss of the projectile particle (mass $m_1$) colliding with a stationary target particle (mas

Vedclass · Accepted Answer

$8/9$

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(B) In a head-on elastic collision,the fractional energy loss of the projectile particle (mass $m_1$) colliding with a stationary target particle (mass $m_2$) is given by the formula:$\frac{\Delta K}{K} = 1 - \left(\frac{m_1 - m_2}{m_1 + m_2}ight)^2$Here,the mass of the neutron $m_1 = 1 	ext{ u}$ and the mass of the deuteron $m_2 = 2 	ext{ u}$.Substituting these values into the formula:$\frac{\Delta K}{K} = 1 - \left(\frac{1 - 2}{1 + 2}ight)^2$$\frac{\Delta K}{K} = 1 - \left(\frac{-1}{3}ight)^2$$\frac{\Delta K}{K} = 1 - \frac{1}{9} = \frac{8}{9}$Thus,the fractional energy loss of the neutron is $8/9$.

$A$ neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is

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