Question

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume all motion is along a straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the of the electron before the collision?

Answer 1

Answer:

$\frac{K_{h}}{K_{i}} = = 2.17 \times 10^{-3}$

Explanation:

As we know that there is no external force on the system of hydrogen atom and electron so we will say momentum is conserved

so we will have

$m_1 v_{1i} = m_1v_{1f} + m_2v_{2f}$

here we know that

$m_1 = m$

$m_2 = 1837m$

now we have

$m v = mv_{1f} + 1837v_{2f}$

$v_{1f} + 1837v_{2f} = v$

also we know that

$v_{2f} - v_{1f} = v$

now we will have

$1838v_{2f} = 2v$

$v_{2f} = \frac{v}{919}$

now we need to find the ratio of kinetic energy of hydrogen atom with initial kinetic energy

so it is given as

$\frac{K_{h}}{K_{i}} = \frac{\frac{1}{2}(1837m)(\frac{v}{919})^2}{\frac{1}{2}mv^2}$

$\frac{K_{h}}{K_{i}} =\frac{1837}{919^2}$

$\frac{K_{h}}{K_{i}} = = 2.17 \times 10^{-3}$