Determine the maximum potential difference that may be applied...

For characteristic X-rays to be emitted, the incident electrons must possess sufficient kinetic energy to knock out an electron from the respective inner shell of the target atom.

From the given data, the energy required to create a vacancy in the $K$-shell (binding energy of the $K$-shell) is $69.5 \text{ keV}$, and the energy required to create a vacancy in the $L$-shell is $11.3 \text{ keV}$.

To avoid emitting any characteristic $K$ X-rays, the energy of the incident electrons must be less than $69.5 \text{ keV}$.
To avoid emitting any characteristic $L$ X-rays, the energy of the incident electrons must be less than $11.3 \text{ keV}$.

In order to emit neither $K$ nor $L$ characteristic X-rays, the maximum kinetic energy of the incident electrons must not exceed the binding energy of the $L$-shell.

Thus, the maximum kinetic energy is E_{\text\max} = 11.3 \text{ keV}.

Since the kinetic energy of an electron accelerated through a potential difference $V$ is given by $E = eV$, the maximum potential difference that can be applied is:
$V = \dfrac{11.3 \text{ keV}}{e} = 11.3 \text{ kV}$