A graph is plotted showing the kinetic energy ( K ) of an electron...

The electrostatic force between the nucleus (charge $Ze$) and the electron (charge $e$) provides the centripetal force:

$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze^2}{r^2}$

Multiplying both sides by $r$, we get:

$mv^2 = \dfrac{1}{4\pi\epsilon_0}\dfrac{Ze^2}{r}$

The kinetic energy $K$ of the electron is:

$K = \dfrac{1}{2}mv^2 = \dfrac{1}{8\pi\epsilon_0}\dfrac{Ze^2}{r}$

This can be written as:

$K = \left(\dfrac{Ze^2}{8\pi\epsilon_0}\right) \times \dfrac{1}{r}$

This is a linear equation of the form $y = mx$, where $y = K$ and $x = \dfrac{1}{r}$.

Thus, the graph of $K$ versus $\dfrac{1}{r}$ is a straight line passing through the origin.

The slope of the graph is $m = \dfrac{Ze^2}{8\pi\epsilon_0}$, which is directly proportional to the atomic number $Z$.

For Hydrogen, $Z = 1$. For He$^+$, $Z = 2$.

Therefore, the slope for He$^+$ is exactly twice the slope for H.

Answer: The graphs are straight lines passing through the origin, and the slope for He$^+$ is twice the slope for H.