## College Physics (4th Edition)

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