If $K_1$ and $K_2$ are the maximum kinetic energies of photo electrons emitted when lights of wavelength $λ_1$ and $λ_2$ respectively incident on a metallic surface and $λ_1 = 3λ_2$. Then -
Answer & explanation
Correct answer: option 2
$K_1=\frac{hc}{λ_1}-\phi$
$K_2=\frac{hc}{λ_2}-\phi$
$\frac{K_2}{K_1}=\frac{(\frac{3hc}{λ_1}-\phi)}{(\frac{hc}{λ_1}-\phi)}=\frac{3(\frac{hc}{λ_1}-\frac{\phi}{3})}{(\frac{hc}{λ_1}-\phi)}$
$\frac{K_2}{K_1}=>3$