Answer
$\boxed{\lambda_{min}=17.89\;pm}$
Work Step by Step
The shortest wavelength in the spectrum is the cutoff wavelength , which is emitted when an incident electron loses its full kinetic energy $K_0$ in a single collision:
$\lambda_{min}=\frac{hc}{K_0}$ .............................$(1)$
Given: $K_0=69.5\;keV=69.5\times10^3\times1.6\times10^{-19}\;J=1.112\times10^{-14}\;J$
Substituting, the given values of $K_0$ in Eq. $1$, we obtain
$\lambda_{min}=\frac{6.63\times 10^{-34}\times3\times 10^{8}}{1.112\times 10^{-14}}\;m$
or, $\lambda_{min}\approx1.789\times 10^{-11}\;m$
or, $\boxed{\lambda_{min}=17.89\;pm}$
