Question

How fast does a proton have to be moving in order to have the same de Broglie wavelength as an electron that is moving at $3.90\times {{10}^{6}}\,m{{s}^{-1}}$ ?

Answer 1

Hint: The de Broglie equation is used to define the wave properties of matter. The electrons possess dual nature property. De Broglie derived an equation related to mass of smaller particles and wavelength. De Broglie suggested that any particle that is moving will be associated with the wave character.

Formula used:$\lambda =\dfrac{h}{mv}$
where, $\lambda $ is the wavelength, $h$ is the Planck’s constant, that is, $6.626\times {{10}^{-34}}J\,s$ , $v$ is the velocity and $m$ is the mass.

Complete step-by-step solution:De Broglie proposed a relation between momentum of a particle and velocity with the wavelength. In this question, it is given that the velocity of electron is $3.90\times {{10}^{6}}\,m{{s}^{-1}}$
Now, let us see de Broglie’s equation.
$\lambda =\dfrac{h}{mv}$
where, $\lambda $ is the wavelength, $h$ is the Planck’s constant, that is, $6.626\times {{10}^{-34}}J\,s$ , $v$ is the velocity and $m$ is the mass.
Now, let us say \[{{m}_{e}}\] and ${{v}_{e}}$ represent mass and velocity of electrons, respectively.
${{m}_{p}}$ and ${{v}_{p}}$ represent mass and velocity of protons, respectively.
Hence, we can write the de Broglie’s equation as:
\[\lambda =\dfrac{h}{{{m}_{e}}{{v}_{e}}}=\dfrac{h}{{{m}_{p}}{{v}_{p}}}\]
The Planck’s constant will be cancelled out in the equation, and it can be rewritten as:
${{v}_{p}}={{v}_{e}}\times \dfrac{{{m}_{e}}}{{{m}_{p}}}$
Mass of electron is $9.109\times {{10}^{-31}}kg$
And mass of proton is $1.673\times {{10}^{-27}}kg$
${{v}_{p}}={{v}_{e}}\times \dfrac{{{m}_{e}}}{{{m}_{p}}}$
Now, substituting the values in the above equation, we get,
${{v}_{p}}=3.90\times {{10}^{6}}\times \dfrac{9.109\times {{10}^{-31}}}{1.673\times {{10}^{-27}}}$
$\Rightarrow {{v}_{p}}=2.12\times {{10}^{3}}\,m{{s}^{-1}}$
Therefore, a proton has to move with a velocity of $2.12\times {{10}^{3}}\,m{{s}^{-1}}$ , which is nearly equal to $2000\,m{{s}^{-1}}$ , in order to have the same de Broglie wavelength as an electron.

Note: The Planck’s constant is a fundamental physical constant that relates a photon energy to its frequency. It is denoted with a symbol $'h'$ .The value of Planck’s constant is $6.626\times {{10}^{-34}}J\,s$
De Broglie’s wavelength is inversely proportional to its momentum. It is used to define the wave properties of matter.