Question

An alpha particle colliding with one of the electrons in a gold atom loses:

• A. Most of its momentum
• B. About $\frac{1}{3}$ rd of its momentum
• C. Little of its energy
• D. Most of its energy

Answer 1

The correct option is A Most of its momentum

man of $\alpha$-particle, $\left(m_1\right)=4amu$

man of an electron $\left(m_2\right)\simeq 5.5\times {10}^{-4}amu$ [ we can take $m_1>>m_2$]

initially, if $\alpha$-particle travels at velocity $u_1$, then carrying a momentum,

$P_1=m_1{u}_1\Rightarrow P_1=4u_1$. collides with electron at rest $(u_2=0)$ we get,

from conversation of linear momentum,

$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2{u}_2}{m_1+m_2}\right)$

as $m_1>>m_2$, we can take, $m_2\simeq 0$ and so,

$V_1\simeq 0\Rightarrow P_1^{\prime }=0$ ( final momentum of $\alpha$-particle)

$\therefore$ Lose of momentum, $=\left(\frac{P_1^{\prime }-P_1}{P_1}\right)\times 100\%\simeq 100\%$.

$\therefore \alpha$-particle loses most of its momentum.