با استفاده از معادلهٔ انیشتین برای فوتوالکتریک داریم:
${K_{\max }} = hf - {w_0} \Rightarrow \left\{ \begin{gathered}
{K_{\max }} = hf - {w_0} \hfill \\
0/6{K_{\max }}h(\frac{3}{4}f) - {w_0} \hfill \\
\end{gathered} \right.$
$ \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{\frac{{8 \times {{10}^{ - 19}}}}{{1/6 \times {{10}^{ - 19}}}} = hf - {w_0}} \\
{\frac{{0/6 \times 8 \times {{10}^{ - 19}}}}{{0/6 \times {{10}^{ - 19}}}} = \frac{3}{4}hf - {w_0}}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{5 = hf - {w_0}} \\
{3 = \frac{3}{4}hf - {w_0}}
\end{array}} \right.$
با حل دو معادله و دو مجهول فوق به دست میآید:
$\left\{ \begin{gathered}
hf = 8ev \hfill \\
{w_0} = 3ev \hfill \\
\end{gathered} \right.$