Question

The allowed energies of a simple atom are 0.0 eV, 3.0 eV, and 4.0 eV. An electron traveling at a speed of 1.3*10^6 m/s collisionally excites the atom.Part A: What is the minimum speed the electron could have after the collision?Part B: What is the maximum speed the electron could have after the collision?

Answer 1

A) $5.34\cdot 10^5 m/s$

The minimum speed of the electron occurs when the electron loses the maximum energy: this occurs when the electron excites the atom from 0.0 eV to 4.0 eV, because in this case the energy given to the atom is maximum.

The energy given by the electron to the atom is equal to the difference between the two energy levels:

$\Delta E= 0.0 eV - 4.0 eV =-4.0 eV = -6.4\cdot 10^{-19}J$

This is equal to the kinetic energy lost by the electron:

$K_f - K_ i = \Delta E\\\frac{1}{2}m(v^2-u^2) = \Delta E$

where

m is the electron's mass

v is the final speed of the electron after the collision

$u=1.3\cdot 10^6 m/s$ is the speed of the electron before the collision

Solving for v, we find

$v=\sqrt{ \frac{2\Delta E}{m}+u^2}=\sqrt{\frac{2 (-6.4\cdot 10^{-19} J)}{9.11\cdot 10^{-31} kg}+(1.3\cdot 10^6 m/s)^2}=5.34\cdot 10^5 m/s$

B) $1.16\cdot 10^6 m/s$

The maximum speed of the electron occurs when the electron loses the minimum amount of energy: this occurs when the electron excites the atom from 3.0 eV to 4.0 eV, because in this case the energy given to the atom is minimum.

The energy given by the electron to the atom is equal to the difference between the two energy levels, so in this case we have:

$\Delta E= 3.0 eV - 4.0 eV =-1.0 eV = -1.6\cdot 10^{-19}J$

And so, this time the final speed of the electron after the collision will be given by:

$v=\sqrt{ \frac{2\Delta E}{m}+u^2}=\sqrt{\frac{2 (-1.6\cdot 10^{-19} J)}{9.11\cdot 10^{-31} kg}+(1.3\cdot 10^6 m/s)^2}=1.16\cdot 10^6 m/s$