Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $$m_p = 1849 m_e$$)

The de Broglie wavelength is given by: $$ \lambda = \frac{h}{p} $$ where $$h$$ is Planck's constant and $$p$$ is the momentum of the particle.

For the proton and electron to have the same de Broglie wavelength, we set $$ \lambda_p = \lambda_e $$ which implies $$ \frac{h}{p_p} = \frac{h}{p_e} $$ and hence $$ p_p = p_e $$. Therefore, the ratio of their momenta is $$ \frac{p_p}{p_e} = \frac{1}{1} $$. The correct answer is 1 : 1.