Question

The radius of an \[\alpha -\text{particle}\] moving in a circle in a constant magnetic field is half of the radius of an electron moving in a circular path in the same field. The de Broglie wavelength of \[\alpha -\text{particle}\] is x times that of the electrons. Find x (an integer).

Answer 1

Hint: To solve this question, we need to understand the basics of the magnetic field. The de Broglie wavelength of a particle is given by the ratio of Planck's constant to its momentum. Radius of path for a particle moving in a magnetic field is given as the ratio of its momentum to the product of its charge and magnetic field it was placed in. An \[\alpha -\text{particle}\] have a charge of +2e where an electron is having only -e.

Formula used:
\[\begin{align}
  & r=\dfrac{P}{Bq} \\
 & \lambda =\dfrac{h}{P} \\
\end{align}\]

Complete step by step answer:
Let us first define de Broglie wavelength. According to the wave-particle duality, the de-Broglie wavelength is defined as the wavelength which is manifested in all the objects in quantum mechanics. The de-Broglie wavelength helps in determining the probability of density of finding the object at a given point of the configuration space.
We should know that the de-Broglie wavelength of a particle is inversely proportional to its momentum.
Since \[r=\dfrac{P}{Bq}\],
\[r\propto \dfrac{P}{q}\]
Here, r is the radius of the circular path in which the particle is moving.
           P is the momentum.
And q is the charge of the particle.
It is given that, ${{r}_{\alpha }}=\dfrac{1}{2}{{r}_{e}}$
Therefore,
\[\begin{align}
  & \dfrac{{{P}_{\alpha }}}{B\left( +2e \right)}=\dfrac{1}{2}\left( \dfrac{{{P}_{e}}}{B\left( -e \right)} \right) \\
 & \Rightarrow {{P}_{\alpha }}=-{{P}_{e}} \\
\end{align}\]
Now, de Broglie wavelength of a particle is given by,
 \[\lambda =\dfrac{h}{P}\]
Where, h is the planck's constant.
      And P is the momentum.
\[\begin{align}
  & \lambda =\dfrac{h}{P} \\
 & \Rightarrow \lambda \propto \dfrac{1}{P} \\
\end{align}\]
Then,
\[\dfrac{{{\lambda }_{\alpha }}}{{{\lambda }_{e}}}=\dfrac{{{P}_{e}}}{{{P}_{\alpha }}}\]
We have already found the relation between momentum of electron and \[\alpha -\text{particle}\] as \[{{P}_{\alpha }}=-{{P}_{e}}\].
Therefore,
\[\begin{align}
  & \dfrac{{{\lambda }_{\alpha }}}{{{\lambda }_{e}}}=\dfrac{{{P}_{e}}}{-{{P}_{e}}} \\
 & \Rightarrow {{\lambda }_{\alpha }}=-{{\lambda }_{e}} \\
\end{align}\]
That means x =-1.
Therefore, the value of x is -1.

Note:
In the answer we have mentioned the concept of magnetic field. When a magnetic field is increasing or decreasing, the force that will be exerted on the electrons is changing. The forces on the electrons are unbalanced as a result, and they will move. When the situation comes that the magnetic field becomes constant, the electrons reach a new point of balance and stop moving.