The ratio of the de-Broglie wavelengths of proton and electron...

The de-Broglie wavelength is given by

$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}} ( \text{As}, {p}^{2} = 2mK )$.

It is given that

$( K{)}_{e} = ( K{)}_{p}$

Thus, the ratio of the wavelengths of the proton and electron is

$\dfrac{{\lambda}_{p}}{{\lambda}_{e}} = \dfrac{\sqrt{2K{m}_{e}}}{\sqrt{2K{m}_{p}}} = \sqrt{\dfrac{{m}_{e}}{{m}_{p}}} = \sqrt{\dfrac{1}{1849}} = \dfrac{1}{43}$.