Answer
$\lambda_{min}=\frac{1240\;pm}{V}$
Work Step by Step
The cutoff wavelength in the continuous x-ray spectrum from any target is the shortest wavelength in the spectrum, which is emitted when an incident electron loses its full kinetic energy $K_0$ in a single collision:
$\lambda_{min}=\frac{hc}{K_0}$
or, $\lambda_{min}=\frac{hc}{eV}$
or, $\lambda_{min}=\frac{(6.626\times 10^{-34}\;J.s)\times(3\times 10^{8})\;m/s}{(1.6\times 10^{-19}\;C)\times V(eV)}\;$
or, $\lambda_{min}=\frac{(6.626\times 10^{-34}\;J.s)\times(3\times 10^{8})\;m/s}{(1.6\times 10^{-19}\;C)\times V(keV)\times 10^{3}}\;m$
or, $\lambda_{min}=\frac{1240\times10^{-12}\;m}{V}$
or, $\lambda_{min}=\frac{1240\;pm}{V}$
