An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?
Answers
Ratio of kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is 2.175 × 10⁻³.
Let take the mass of electron be m.
So according to the question mass of Hydrogen will be 1837 times of m. So, M = 1837 m
Initial velocity of the electron is u.
By applying momentum conservation we know that initial momentum should be equal to final momentum.
P1 = P2
mu = mV1+ MV2
mu = mV1+ 1837 m V2
V1 + 1837 V2 = u -----------(i)
We know that for elastic collision
V2- V1 = u -----------(ii)
From the equation of (i) and (ii) we can find out the value of V1 and V2
V1= - 918 u /919 m/sec
V2= u/919 m/sec
Now we have to find out the ratio of kinetic energy of the hydrogen atom to the electron
Kh/Ke = 0.5 MV₂² / 0.5 mu²
Kh/ke = 1837×m× (u/919)² / 0.5 mu²
Kh/Ke = 2.175 × 10⁻³