Physics - Modern Physics Question with Solution | TestHub

The de-Broglie wavelength is given by 

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

It is given that

 $(K{)}_e=(K{)}_p$

Thus, the ratio of the wavelengths of the proton and electron is 
       $$
      $\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{2K{m}_e}}{\sqrt{2K{m}_p}}=\sqrt{\frac{m_e}{m_p}}=\sqrt{\frac{1}{1849}}=\frac{1}{43}$.