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

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:

& r=\dfrac{P}{Bq} \\\ & \lambda =\dfrac{h}{P} \\\ \end{aligned}$$ **Complete step by step answer:** Let us first define de Broglie wavelength. According to the wave-particle duality, the de-Broglie wavelength is defined as the wavelength which is manifested in all the objects in quantum mechanics. The de-Broglie wavelength helps in determining the probability of density of finding the object at a given point of the configuration space. We should know that the de-Broglie wavelength of a particle is inversely proportional to its momentum. Since $$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{aligned} & \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{aligned}$$ Now, de Broglie wavelength of a particle is given by, $$\lambda =\dfrac{h}{P}$$ Where, h is the planck's constant. And P is the momentum. $$\begin{aligned} & \lambda =\dfrac{h}{P} \\\ & \Rightarrow \lambda \propto \dfrac{1}{P} \\\ \end{aligned}$$ Then, $$\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{aligned} & \dfrac{{{\lambda }_{\alpha }}}{{{\lambda }_{e}}}=\dfrac{{{P}_{e}}}{-{{P}_{e}}} \\\ & \Rightarrow {{\lambda }_{\alpha }}=-{{\lambda }_{e}} \\\ \end{aligned}$$ 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.