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?

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aristocles aristocles · Answer 1 · 2019-10-28T05:12:54+01:00

Answer:

[tex]\frac{K_{h}}{K_{i}} = = 2.17 \times 10^{-3}[/tex]

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

[tex]m_1 v_{1i} = m_1v_{1f} + m_2v_{2f}[/tex]

here we know that

[tex]m_1 = m[/tex]

[tex]m_2 = 1837m[/tex]

now we have

[tex]m v = mv_{1f} + 1837v_{2f}[/tex]

[tex]v_{1f} + 1837v_{2f} = v[/tex]

also we know that

[tex]v_{2f} - v_{1f} = v[/tex]

now we will have

[tex]1838v_{2f} = 2v[/tex]

[tex]v_{2f} = \frac{v}{919}[/tex]

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

so it is given as

[tex]\frac{K_{h}}{K_{i}} = \frac{\frac{1}{2}(1837m)(\frac{v}{919})^2}{\frac{1}{2}mv^2}[/tex]

[tex]\frac{K_{h}}{K_{i}} =\frac{1837}{919^2}[/tex]

[tex]\frac{K_{h}}{K_{i}} = = 2.17 \times 10^{-3}[/tex]