Electron and Hydrogen Collision

Classical Mechanics Level 4

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?

E k = 1 2 m v 2 E_k = \frac{1}{2}mv^2


The answer is 0.0022.

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Zeyad Moustafa by Zeyad Moustafa , 21, USA

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1 solution

Tapas Mazumdar
Mar 29, 2018

Let the speed of the electron before collision be v v and that after the collision the speeds of the electron and the hydrogen atom be v 1 v_1 and v 2 v_2 respectively. Taking the direction of the initial velocity of the electron to be positive and using conservation of linear momentum and Newton's Law of restitution, we get

m v = − m v 1 + 1837 m v 2 , e = 1 = ∣ v 2 + v 1 ∣ ∣ v ∣ = v 1 + v 2 v mv = -mv_1 + 1837m v_2 \quad , \quad e = 1 = \dfrac{|v_2 + v_1|}{|v|} = \dfrac{v_1 + v_2}{v}

Solving the two equations above, we get v 2 = v 919 v_2 = \dfrac{v}{919} .

Required ratio is K H K e = 1 2 ( 1837 m ) ( v 919 ) 2 1 2 m v 2 = 1837 ( 919 ) 2 ≈ 0.0022 \dfrac{K_H}{K_e} = \dfrac{\frac{1}{2} (1837m) {\left(\frac{v}{919}\right)}^2}{\frac{1}{2} mv^2} = \dfrac{1837}{(919)^2} \approx \boxed{0.0022}

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