Explanation:
Cut off wavelength is given by
$\lambda_{\text{min}}=\frac{\text{hc}}{\text{eV}'}$
where h = Planck's constan
c = speed of light
e = charge on an electron
V = potential difference applied to the tube
When potential difference (V) applied to the tube is doubled, cutoff wavelength
$\big({\lambda}'_\text{min}\big)$ is given by
${\lambda}'_\text{min}=\frac{\text{hc}}{\text{e}(2\text{V})}$
$\Rightarrow\lambda'_\text{min}=\frac{\lambda_\text{min}}{2}$
Cuttoff wavelength does not depend on the separation between the filament and the target.
Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.
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