The energy of a photon with frequency $\nu$ is $h \nu$ , where $h = 2 \pi \hbar$ is Planck's constant. Assuming all of this energy is transferred to a single a electron, the electron will have energy $h \nu - \phi$ after being ejected from the metal. The potential difference required to decelerate such an electron until it stops is $V = (h \nu - \phi)/e$ . So, the slope of the line is $\boxed{h/e}$ . Therefore, answer (B) is correct.

Solution to 2008 Problem 41