Question

If velocity of a particle is three times that of electron and ratio of de Broglie wavelength of particle to that of electron is $1.814\times {{10}^{-4}}$. The particle will be:
A) neutron
B) deuteron
C) alpha
D) tritium

Answer 1

Hint: The DE Broglie’s wavelength depends on the Planck’s constant and the momentum of the particle. As momentum is the product of mass and the velocity of the particle, the DE Broglie’s wavelength changes as the velocity and mass changes. Velocity of the unknown particle is given in term so the velocity of the electron and ratio is given. We can easily find the mass of an unknown particle.
Formula used:
$\lambda =\dfrac{h}{mv}$

Complete answer:
Let us write down the given terms and quantities to us,
$\begin{align}
  & {{v}_{p}}=3{{v}_{e}} \\
 & \Rightarrow \dfrac{{{\lambda }_{p}}}{{{\lambda }_{e}}}=1.814\times {{10}^{-4}} \\
\end{align}$
Now, we know the de Broglie’s wavelength is equal to,
$\lambda =\dfrac{h}{mv}$
Let us write down the DE Broglie’s wavelength of both particles and divide them,
\[\begin{align}
  & {{\lambda }_{p}}=\dfrac{h}{{{m}_{p}}{{v}_{p}}} \\
 & \Rightarrow {{\lambda }_{e}}=\dfrac{h}{{{m}_{e}}{{v}_{e}}} \\
 & \Rightarrow 1.814\times {{10}^{-4}}=\dfrac{{{m}_{e}}{{v}_{e}}}{{{m}_{p}}{{v}_{p}}} \\
 & \Rightarrow {{m}_{p}}=\dfrac{9.1\times {{10}^{-31}}}{1.814\times {{10}^{-4}}\times 3} \\
 & \Rightarrow {{m}_{p}}=1.672\times {{10}^{-27}}kg \\
\end{align}\]
If we observe the mass of the unknown particle, the mass of the unknown particle is equal to the mass of the neutron.

So, the correct answer is “Option A”.

Additional Information:
According to wave particle duality, the DE Broglie wavelength is available and manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. The Broglie’s wavelength of a particle is inversely proportional to its momentum. Matter waves add a central part of the theory of quantum mechanics all matter exhibits wavelike behaviour. A beam of electrons can be diffracted just like a beam of light or a water wave. The wavelength in most cases is however too small to have practical impact on day to day activities. Matter waves are not relevant for objects with size such as cricket balls and people.

Note:
Matter waves are used to explain that every object exhibits wave nature and the waves of the object are matter waves. Light also exhibits both particle and wave nature. Similar to light, all objects have dual behaviour.