Chemistry Question from JEE Main 2026 · 23 January, Shift 2

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 $6 \, \text{eV}$, the kinetic energy of liberated electrons of $M_A$ : $M_B$ is in the ratio of $2.642 : 1$. The work functions (in $\text{eV}$) of $M_A$ and $M_B$ are respectively.