Kinetic Energy Ratio At Equal De Broglie Wavelength

Equal de Broglie wavelengths mean equal momenta, because $\lambda =\frac{h}{p}$ depends only on momentum, so $p_e=p_p$. For a non-relativistic particle the kinetic energy in terms of momentum is $K=\frac{p^2}{2m}$, which shows that at fixed momentum the kinetic energy is inversely proportional to mass. Hence $\frac{K_e}{K_p}=\frac{m_p}{m_e}$. Using $m_p=1.67\times 10^{-27}\text{kg}$ and $m_e=9.1\times 10^{-31}\text{kg}$, the ratio is about $1836:1$. The option $1:1836$ inverts the mass dependence by wrongly putting the lighter mass on top. The option $\sqrt{1836}:1$ would apply to a speed ratio, not an energy ratio. The option $1:1$ ignores the mass difference entirely. The physical insight is that for the same wavelength the much lighter electron must carry vastly more kinetic energy than the heavy proton, since each has identical momentum but kinetic energy scales as the inverse of mass. This subtle distinction between momentum-matched and energy-matched comparisons is a recurring JEE theme. A consistency check confirms the lighter particle holds the larger kinetic energy, exactly as $K\propto 1/m$ predicts. A complementary scenario worth contrasting is one where the two particles instead share the same kinetic energy; then the heavier proton would carry more momentum and hence have the shorter wavelength, reversing the intuition. Recognising whether a problem fixes wavelength, momentum, speed, or energy is the single most important step, since each constraint produces a different dependence on mass.