Question

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

Answer 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:
$\lambda ={\displaystyle \frac{h}{mv}}$
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 $m_p=1836m_e$

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

 The de Broglie wavelength for proton is calculated as ${\lambda }_{proton}={\displaystyle \frac{h}{m_{p}v}}\to (ii)$

Now, divide the equation (i) by (ii) we get,
$$\begin{gathered} {\displaystyle \frac{{\lambda }_{electron}}{{\lambda }_{proton}}}={\displaystyle \frac{{\displaystyle \frac{h}{m_{e}v}}}{{\displaystyle \frac{h}{m_{p}v}}}} \\ {\displaystyle \frac{{\lambda }_{electron}}{{\lambda }_{proton}}}={\displaystyle \frac{m_p}{m_e}} \end{gathered}$$

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

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

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.