Answer
(a) electron
(b) $1836$
Work Step by Step
(a) We know that mass of the electron is given as
$m_e=\frac{h^2}{2E_e\lambda_e^2}$
Similarly, the mass of the proton is given as
$m_p=\frac{h^2}{2E_p\lambda_p^2}$
As $m_p\gt m_e$
therefore $\frac{h^2}{2E_p\lambda_p^2}\gt \frac{h^2}{2E_e\lambda_e^2}$
$\implies \frac{1}{E_p}\gt \frac{1}{E_e}$
$\implies E_p\gt E_e$
Thus, for the same de Broglie wavelength of the electron and proton, the electron has greater kinetic energy than the proton.
(b) We know that
$\frac{E_e}{E_p}=\frac{m_p}{m_e}$
We plug in the known values to obtain:
$\frac{E_e}{E_p}=\frac{1.673\times 10^{-27}Kg}{9.11\times 10^{-31}Kg}$
$\frac{E_e}{E_p}=1836$
