Answer
$\boxed{V_{min}=69.5\;kV}$
Work Step by Step
To produce the characteristic x-ray spectrum $K_{\alpha}$ and $K_{\beta}$ of tungsten, an electron has to be knocked out from the deep-lying $K$ shell $(n=1)$. The minimum energy required to remove an electron from $K$ shell is just the energy of the shell.
It is given that the $K$ energy level for tungsten have the energy $69.5\;keV$. Therefore, the minimum value of the accelerating potential $(V_{min})$ that will permit the production of the characteristic $K_{\alpha}$ and $K_{\beta}$ lines of tungsten is: $\boxed{V_{min}=69.5\;kV}$
