Question

consider a metal surface exposed to light of frequency 600nm the maximum energy of photoelectrons doubles when light of wavelength 400 nm is used find the work function of the metal

Answer 1

$$\begin{gathered} DearStudent, \\ \frac{hc}{{\lambda }_1}-{\phi }_o=k-----\left(1\right) \\ \frac{hc}{{\lambda }_2}-{\phi }_o=2k-----\left(2\right) \\ from\left(i\right)and\left(ii\right) \\ \frac{hc}{{\lambda }_2}-{\phi }_o=2\left(\frac{hc}{{\lambda }_1}-{\phi }_o\right) \\ \frac{hc}{{\lambda }_2}-{\phi }_o=2\frac{hc}{{\lambda }_1}-2{\phi }_o \\ {\phi }_o=hc\left[\frac{2}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right] \\ {\phi }_o=1230\left[\frac{2}{600}-\frac{1}{400}\right] \\ Soweget{\phi }_o=1.02ev \end{gathered}$$
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