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

`(K_(h))/(K_(i)) = = 2.17 * 10^(-3)`

Step-by-step 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) = (v)/(919)`

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

so it is given as

`(K_(h))/(K_(i)) = ((1)/(2)(1837m)((v)/(919))^2)/((1)/(2)mv^2)`

`(K_(h))/(K_(i)) =(1837)/(919^2)`

`(K_(h))/(K_(i)) = = 2.17 * 10^(-3)`