In a photoelectric experiment using a sodium surface, you find a stopping potential of $$1.85 \mathrm{V}$$ for a wavelength of $$300 \mathrm{nm}$$ and a stopping potential of $$0.820 \mathrm{V}$$ for a wavelength of $$400 \mathrm{nm} .$$ From these data find (a) a value for the Planck constant, (b) the work function $$\Phi$$ for sodium, and (c) the cutoff wavelength $$\lambda_{0}$$ for sodium.
(a) For the first and second case (labeled 1 and 2) we have
$$e V_{01}=h c / \lambda_{1}-\Phi, \quad e V_{02}=h c / \lambda_{2}-\Phi\\$$
from which $$h$$ and $$\Phi$$ can be determined.
Thus,
$$h=\dfrac{e\left(V_{1}-V_{2}\right)}{c\left(\lambda_{1}^{-1}-\lambda_{2}^{-1}\right)}=\dfrac{1.85 \mathrm{eV}-0.820 \mathrm{eV}}{\left(3.00 \times 10^{17} \mathrm{nm} / \mathrm{s}\right)\left[(300 \mathrm{nm})^{-1}-(400 \mathrm{nm})^{-1}\right]}=4.12 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s}\\$$
(b) The work function is
$$\Phi=\dfrac{3\left(V_{2} \lambda_{2}-V_{1} \lambda_{1}\right)}{\lambda_{1}-\lambda_{2}}=\dfrac{(0.820 \mathrm{eV})(400 \mathrm{nm})-(1.85 \mathrm{eV})(300 \mathrm{nm})}{300 \mathrm{nm}-400 \mathrm{nm}}=2.27 \mathrm{eV}\\$$
(c) Let $$\Phi=h c / \lambda_{\max }$$ to obtain
$$\lambda_{\max }=\dfrac{h c}{\Phi}=\dfrac{1240 \mathrm{eV} \cdot \mathrm{nm}}{2.27 \mathrm{eV}}=545 \mathrm{nm}\\$$