de-Broglie wavelength of a proton and an electron is same. The ratio of kinetic energy of electron to that of proton is

The de-Broglie wavelength ($\lambda$) of a particle is related to its momentum ($p$) by the relation:
$\lambda =\frac{h}{p}$
where $h$ is Planck's constant.

The momentum of a particle can be written in terms of its mass ($m$) and kinetic energy ($K$) as:
$p=\sqrt{2mK}$
Substituting this into the de-Broglie wavelength formula gives:
$\lambda =\frac{h}{\sqrt{2mK}}$

Let ${\lambda }_e$ be the de-Broglie wavelength of the electron and ${\lambda }_p$ be the de-Broglie wavelength of the proton. According to the question, their wavelengths are equal:
${\lambda }_e={\lambda }_p$
This implies:
$\frac{h}{\sqrt{2m_e{K}_e}}=\frac{h}{\sqrt{2m_p{K}_p}}$
where $m_e$ and $K_e$ are the mass and kinetic energy of the electron, and $m_p$ and $K_p$ are the mass and kinetic energy of the proton.

Squaring both sides and simplifying, we get:
$m_e{K}_e=m_p{K}_p$
Rearranging the terms to find the ratio of the kinetic energy of the electron to that of the proton:
$\frac{K_e}{K_p}=\frac{m_p}{m_e}$

The mass of a proton ($m_p$) is approximately 1835 times the mass of an electron ($m_e$):
$\frac{m_p}{m_e}\approx 1835$
Therefore, the ratio of the kinetic energy of the electron to that of the proton is:
$\frac{K_e}{K_p}=1835$