A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume $$h = 6.63 \times 10^{-34} \text{ J s}$$, $$m_e = 9.0 \times 10^{-31} \text{ kg}$$ and $$m_p = 1836 \; m_e$$)

A proton and electron have the same de Broglie wavelength. Find the ratio of their kinetic energies.

$$\lambda = \frac{h}{p} \implies p = \frac{h}{\lambda}$$

Since both have the same $$\lambda$$, they have the same momentum $$p$$.

$$KE = \frac{p^2}{2m}$$

Since $$p$$ is the same for both:

$$\frac{KE_p}{KE_e} = \frac{p^2/(2m_p)}{p^2/(2m_e)} = \frac{m_e}{m_p} = \frac{m_e}{1836 \, m_e} = \frac{1}{1836}$$

$$KE_p : KE_e = 1 : 1836$$.

The correct answer is Option (4): $$1 : 1836$$.