If the potential difference applied to the tube is doubled and the...

The cutoff wavelength of X-rays produced in an X-ray tube is given by the relation \lambda_\min = \dfrac{hc}{eV}.

The cutoff wavelength depends only on the accelerating potential difference $V$ and is independent of the separation between the filament and the target.

When the potential difference is doubled, $V' = 2V$.

The new cutoff wavelength becomes \lambda'_\min = \dfrac{hc}{e(2V)} = \dfrac{\lambda_\min}{2}.

Therefore, the cutoff wavelength will be halved.