Chapter 4 Early Quantum Theory - Problems - Page 133: 30

$5.97\cdot 10^{14}s^{-1}$

We have \begin{align*}\frac{m_ev^2}{2}&=E_k=h\left(\frac{c}{39\cdot 10^{-8}m}-\nu_0\right)\\\Longrightarrow \nu_0&=\frac{3\cdot 10^8 ms^{-1}}{39\cdot 10^{-8}m}-\frac{9.1\cdot 10^{-31}kg\cdot 25\cdot 10^{10}m^2s^{-2}}{2\cdot 6.63\cdot 10^{-34}m^2kgs^{-1}}\\& =7.69\cdot 10^{14}Hz-1.715\cdot 10^{15}Hz=5.97\cdot 10^{14}s^{-1}\end{align*}