Answer to Question #279865 in Classical Mechanics for Vishu

Question

Show that energy transferred by energetic electron to helium atom by elastic collision is less than 0.1%.

Answer 1

$\text{Take the frame of reference at a fixed helium nucleus}$

$m_1 - \text{helium atom }$

$m_2 - \text{electron}$

$v_1 = 0$

$EK = \frac{m_2v_2^2}{2}$

$m_2\vec{v_2}=m_1\vec{u_1} +m_2\vec{u_2}$

$m_1\vec{u_1} = m_2\vec{v_2}-m_2\vec{u_2}$

$max (\vec u_{1})= \frac{2\vec v_2m_2}{m_1}$

$\Delta EK_{m_1}= \frac{m_1u_1^2}{2}$

$max(\Delta EK_{m_1})= \frac{2m_2^2v_2^2}{m_1}$

$\text{maxpercentage}= \frac{max(\Delta EK_{m_1})}{EK}*100= \frac{4m_2}{m_1}*100$

$\text{The helium nucleus - is made up of two protons and two neutrons.}$

$\text{The mass of a proton is 1836 heavier than the mass of an electron}$

$\text{maxpercentage}= = \frac{4m_2}{4*1836*m_2}*100= 0.054 \%$

$\text{maxpercentage}<0.1 \%$

$\text{Proven}$