contestada

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?.

Respuesta :

2.17 x [tex]{10}^{-3}[/tex] is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision.

What is kinetic energy?

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_1v_{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_1v_{1i} =mv_{1f} +1837mv_{2f}[/tex]

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

Also we know that

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

Now we will have

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

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

so it is given as

[tex]\frac{K_h}{K_i} =\frac{1837}{919 X 919}[/tex]

[tex]\frac{K_h}{K_i} = 2.17 X {10}^-3[/tex]

Hence,  2.17 x[tex]{10}^{-3}[/tex] is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision.

Learn more about the kinetic energy here:

https://brainly.com/question/22174271

#SPJ1

Otras preguntas