Question

The Frank Hertz experiment involved shooting electrons into a low density gas of mercury atoms and observing discrete amounts of kinetic energy loss by the electrons. Suppose that instead the similar experiment is done with a very cold gas of atomic hydrogen, so that all of the hydrogen atoms are initially in ground state. If the kinetic energy of an electron is 11.6 eV just before it collides with a hydrogen atom. How much kinetic energy will the electron have just after it collides with and excites the hydrogen atom?

Short answer

The kinetic energy of electron after collision is$1.4eV$ .

Step-by-step solution

Identification of given data

The kinetic energy of electron just before collision is$K_i=11.6\text{eV}$

Conceptual Explanation

The electron loses it kinetic energy just after collision with atomic hydrogen and kinetic energy of electron just after collision will be equal to the difference between kinetic energy before collision and energy of hydrogen atom in first excited state.

Determination of kinetic energy of electron after collision

The kinetic energy of electron just after collision is given as:

$K_f=K_i-E_e$

Here, $E_e$ is the energy of hydrogen in first excited state and its value is $10.2eV$

Substitute all the values in the above equation.

$$\begin{gathered} K_f=11.6\text{eV}-10.2\text{eV} \\ {K}_f=1.4\text{eV} \end{gathered}$$

Therefore, the kinetic energy of electron just after collision is $1.4eV$.