Answer
See answers.
Work Step by Step
Use equation 27–4, E = hf, and recall that $\lambda f = c$.
The energy of the photon with the shortest wavelength, i.e., the cutoff wavelength, equals the maximum KE of the electron, eV. Assume that the voltage V is measured in volts.
$$E=hf=\frac{hc}{\lambda_0}=eV$$
Solve for the cutoff wavelength.
$$\lambda_0=\frac{hc}{eV} $$
$$\lambda_0=\frac{(6.626\times10^{-34}J \cdot s)(2.998\times10^8m/s)(10^9\;nm/m)}{(1.60\times10^{-19}C)(Voltage)}$$
$$\lambda_0=\frac{1240nm}{Voltage} $$
