Question 9.121: An electron undergoes a one-dimensional elastic collision wi......

We use m_1 for the mass of the electron and m_2 = 1840m_1 for the mass of the hydrogen atom. Using Eq. 9-68,

v_{2 f}=\frac{2 m_1}{m_1+m_2}v_{1 i}\text{.}       (9-68)

v_{2 f}=\frac{2 m_1}{m_1+1840 m_1}v_{1 i}=\frac{2}{1841}v_{1 i}

we compute the final kinetic energy of the hydrogen atom:

K_{2 f}=\frac{1}{2}b 1840 m_1 g \lceil \frac{2 v_{1 i}}{1841}|^2=\frac{(1840)(4)}{1841^2}\lceil\frac{1}{2}b 1840 m_1 g v_{1 i}^2|

so we find the fraction to be b1840gb4g/ 1841² ≈ 2 2 × 10 ^{-3} , or 0.22%.