Answer
a) The de Broglie wavelength of the electron is longer than the proton.
b) $\frac{\lambda_e}{\lambda_p}=1.83\times 10^3$
Work Step by Step
(a) As $\lambda=\frac{h}{p}=\frac{h}{mv}$
$\implies \lambda \propto \frac{1}{m}$
Thus, wavelength and mass are inversely proportional. We know that the mass of an electron is smaller than that of a proton, hence the de Broglie wavelength of the electron is longer than the proton.
(b) We can calculate the required ratio as
we know that $\lambda\propto \frac{1}{m}$
$\implies \frac{\lambda_e}{\lambda_p}=\frac{m_p}{m_e}$
We plug in the known values to obtain:
$\frac{\lambda_e}{\lambda_p}=\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}$
$\frac{\lambda_e}{\lambda_p}=1.83\times 10^3$
