List of top Chemistry Questions on Dual nature of matter asked in JEE Main

The work functions of two metals ($M_A$ and $M_B$) are in the 1 : 2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of 2.642 : 1. The work functions (in eV) of $M_A$ and $M_B$ are respectively.

Wavelengths of two photons are given as \(\lambda_1 = 3000 \, \text{Å}\) and \(\lambda_2 = 6000 \, \text{Å}\) respectively. Calculate the ratio of their energies \(\left( \frac{E_1}{E_2} \right)\).