An electron of mass ${m_e}$ and a proton of mass ${m_p} = 1836{m_e}$ are moving with the same speed. The ratio of their de Broglie wavelength $\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}}$ will be:
(A) $918$
(B) $1836$
(C) $\dfrac{1}{{1836}}$
(D) $1$

Hint: In order to solve this question, we will first calculate the de Broglie wavelength for an electron of given mass and velocity and then de Broglie wavelength for a proton and then we will find the required ratio of their wavelengths.

Formula Used:
The de Broglie wavelength is calculated using the formula:
$\lambda = \dfrac{h}{{mv}}$
where
h-The Planck’s constant
m-The mass of a particle and
v - The velocity of the particle

Complete answer:
We have given that, an electron and proton are moving with the same speed, let us assume their velocity is $v$ and the relation between the mass of the electron and proton is given as ${m_p} = 1836{m_e}$

Now, the de Broglie wavelength for an electron is calculated using the formula
$\lambda = \dfrac{h}{{mv}}$ we get,
${\lambda _{electron}} = \dfrac{h}{{{m_e}v}} \to (i)$

 The de Broglie wavelength for proton is calculated as ${\lambda _{proton}} = \dfrac{h}{{{m_p}v}} \to (ii)$

Now, divide the equation (i) by (ii) we get,
$
  \dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{\dfrac{h}{{{m_e}v}}}}{{\dfrac{h}{{{m_p}v}}}} \\
  \dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}} \\
 $

Since, we have given that ${m_p} = 1836{m_e}$ or by rearranging it we have $\dfrac{{{m_p}}}{{{m_e}}} = 1836$

Substituting the above value in the equation we get:
$\dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = \dfrac{{{m_p}}}{{{m_e}}}$
$\Rightarrow \dfrac{{{\lambda _{electron}}}}{{{\lambda _{proton}}}} = 1836$

Therefore, the ratio of the wavelength of electron and proton is 1836.

Hence, the correct option is (B) 1836

Note: It should be remembered that de Broglie wavelength is the wavelength associated with all the particles however larger bodies having larger mass show a very negligible amount of wave nature whereas for elementary particles this effect is very noticeable.