Answer
$\Delta V = 4970~V$
Work Step by Step
We can find the minimum potential difference $\Delta V$ applied to the tube:
$q~\Delta V = \frac{hc}{\lambda}$
$\Delta V = \frac{hc}{q~\lambda}$
$\Delta V = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.6\times 10^{-19}~C)(0.250\times 10^{-9}~m)}$
$\Delta V = 4970~V$
