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

Question 32368

One needs to use the laws of conservation of linear momentum and energy.

First one gives:

1) $m_{e}v = m_{e}v^{\prime} + M_{H}v^{\prime \prime}$

, where $m_e$ is the mass of electron, $\nu$ is the initial electron velocity, $\nu'$ is the final electron velocity, $M_H$ is the mass of hydrogen atom and $\nu''$ is the velocity of Hydrogen atom after collision. Also, $\frac{M_H}{m_e} = 1837$ (according to given conditions).

Law of conservation of energy gives:

2) $m_{e}v^{2} = m_{e}v^{\prime 2} + M_{H}v^{\prime \prime 2}$

From equation 1), obtain $v' = v - \frac{M_H}{m_e} v''$ .

Plugging latter expression into 2):

$$m _ {e} v ^ {2} = m _ {e} \left(v - \frac {M _ {H}}{m _ {e}} v ^ {\prime \prime}\right) ^ {2} + M _ {H} v ^ {\prime \prime 2}.$$

Opening brackets in latter expression, obtain $0 = -2 \nu \nu'' M_H + \frac{M_H^2}{m_e} \nu''^2 + M_H \nu''^2$ . Dividing by $m_e \nu''$ , get $2 \nu \frac{M_H}{m_e} = \nu'' \left( \frac{M_H^2}{m_e^2} + \frac{M_H}{m_e} \right)$ , from which obtain connection between $\nu''$ and $\nu$ (using $\frac{M_H}{m_e} = 1837$ ): $\nu'' = \frac{2 \cdot 1837}{(1837^2 + 1837)} \nu = 1.09 \cdot 10^{-3} \nu$ .

The ratio of kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is:

$$k = \frac {\frac {M _ {H} v ^ {\prime 2}}{2}}{\frac {m _ {e} v ^ {2}}{2}} = \frac {M _ {H} v ^ {\prime 2}}{m _ {e} v ^ {2}} = 1 8 3 7 \cdot (1. 0 9 \cdot 1 0 ^ {- 3}) ^ {2} = 0. 0 0 2 2.$$