Problem 28.9

28.9. a)
We know that the energy given to a photon leaving a metal is given by
$hf = e V_{s,1} + \phi_{1}$
$hf = e V_{s,2} + \phi_{2}$
So, we have that
V_s,1 = V_s,2 + 1.48 V
Or otherwise stated, $\phi_{2} - \phi_{1} = 1.48 eV$
From the second condition, $h f_{1} = \phi_{1} = 0.60 h f_{2} = 0.60 \phi_{2}$
Or put another way,
$\phi_{2} - 0.60 \phi_{2} = 1.48 eV$
Solving and then plugging back in to find $\phi_{1}$, we find
$\phi_{2} = 3.70 eV$
$\phi_{1} = 2.22 eV$