Ratio Of Photoelectron Speeds For Two Wavelengths

The maximum kinetic energy for each wavelength follows Einstein's equation $K_{max}=\frac{hc}{\lambda }-{\varphi }_0$, and since $K_{max}=\frac{1}{2}m{v}_{max}^2$, the speed ratio is $\frac{v_1}{v_2}=\sqrt{\frac{K_1}{K_2}}$. For ${\lambda }_1=310\text{nm}$ the photon energy is $\frac{1240}{310}=4.0\text{eV}$, so $K_1=4.0-1.0=3.0\text{eV}$. For ${\lambda }_2=620\text{nm}$ the photon energy is $\frac{1240}{620}=2.0\text{eV}$, so $K_2=2.0-1.0=1.0\text{eV}$. Hence $\frac{v_1}{v_2}=\sqrt{\frac{3.0}{1.0}}=\sqrt{3}$. The option $3:1$ wrongly equates the speed ratio with the kinetic-energy ratio, forgetting the square root. The option $1:\sqrt{3}$ inverts the result. The option $\sqrt{2}:1$ would arise if the work function were ignored. The key subtlety is that speeds scale with the square root of kinetic energy, not of photon energy, so the work function must be subtracted before taking the ratio. This careful two-step reasoning, drawn straight from the photoelectric equation, is exactly the kind of layered analysis JEE Advanced rewards. A check confirms the shorter wavelength yields the faster electrons, so the ratio correctly exceeds one. It is illuminating to note that had the work function been zero, the kinetic energy would scale directly as photon energy and hence inversely as wavelength, giving a much simpler speed ratio. The presence of the finite work function breaks this simple proportionality and is exactly why the subtraction must be performed before forming any ratio, a trap that catches many students.