Physics - Modern Physics Question with Solution | TestHub

Maximum kinetic energies of photo electrons for light of wavelength${\mathrm{K}}_1=\frac{hc}{{\lambda }_1}-f$,

Maximum kinetic energies of photo electrons for light of wavelength${\mathrm{K}}_2=\frac{hc}{{\lambda }_2}-f$

Also given,${\lambda }_1=3{\lambda }_2$

$\frac{K_2}{K_1}=\frac{\left(\frac{3hc}{{\lambda }_1}-\varphi \right)}{\left(\frac{hc}{{\lambda }_1}-\varphi \right)}=\frac{3\left(\frac{hc}{{\lambda }_1}-\frac{\varphi }{3}\right)}{\left(\frac{hc}{{\lambda }_1}-\varphi \right)}$

Therefore$\frac{K_2}{K_1}>3$