Question

An $\alpha$ particle and a proton are accelerated from rest by the same potential. Find the ratio of their de-Broglie wavelength.

Answer 1

- Hint: First, we will find out the kinetic energies of both the $\alpha$ and the proton respectively with the formula $KE={\displaystyle \frac{1}{2}}m{v}^2$. Then we will solve the equations further and find out the equation for momentums of both the particles. Refer to the solution below.

Formula used: $KE={\displaystyle \frac{1}{2}}m{v}^2$, $p=mv$, $\lambda ={\displaystyle \frac{h}{p}}$.

Complete step-by-step solution:
Let the mass of $\alpha$particle be $m_{\alpha }$.
Let the mass of the proton be $m_p$.
Let the velocity of $\alpha$ particle be $v_{\alpha }$.
Let the velocity of the proton be $v_p$.
Now, as we know the formula for kinetic energy is $KE={\displaystyle \frac{1}{2}}m{v}^2$.
Kinetic energy of $\alpha$ particle will be-
$\Rightarrow K{E}_{\alpha }={\displaystyle \frac{1}{2}}{m}_{\alpha }{v_{\alpha }}^2$
Multiplying the numerator and denominator by $m_{\alpha }$, we get-
$\Rightarrow K{E}_{\alpha }={{\displaystyle \frac{\left(m_{\alpha }{v}_{\alpha }\right)}{2m_{\alpha }}}}^2$
As we know that the formula for momentum is $p=mv$. So, from the above equation we get-
$$\begin{gathered} \Rightarrow K{E}_{\alpha }={{\displaystyle \frac{\left(p_{\alpha }\right)}{2m_{\alpha }}}}^2 \\ \Rightarrow {p_{\alpha }}^2=2m_{\alpha }{\left(KE\right)}_{\alpha } \\ \Rightarrow p_{\alpha }=\sqrt{2m_{\alpha }{\left(KE\right)}_{\alpha }} \end{gathered}$$
Kinetic energy of proton will be-
$\Rightarrow K{E}_p={\displaystyle \frac{1}{2}}{m}_p{v_p}^2$
Multiplying the numerator and denominator by $m_p$, we get-
$\Rightarrow K{E}_p={{\displaystyle \frac{\left(m_p{v}_p\right)}{2m_p}}}^2$
As we know that the formula for momentum is $p=mv$. So, from the above equation we get-
$$\begin{gathered} \Rightarrow K{E}_p={{\displaystyle \frac{\left(p_p\right)}{2m_p}}}^2 \\ \Rightarrow {p_p}^2=2m_p{\left(KE\right)}_p \\ \Rightarrow p_p=\sqrt{2m_p{\left(KE\right)}_p} \end{gathered}$$
Now, the work done in accelerating the proton and the $\alpha$ particle will be equal to the kinetic energy acquired. As we know, $W=qV$. Potential difference is the same in both cases. So-
Kinetic energy of $\alpha$ particle in terms of charge and potential difference-
$\Rightarrow K{E}_{\alpha }=q_{\alpha }V$
Kinetic energy of $p$ particle in terms of charge and potential difference-
$\Rightarrow K{E}_p=q_{p}V$
Putting the above values of kinetic energy into the values of momentums, we get-
For $\alpha$ particle-
$$\begin{gathered} \Rightarrow p_{\alpha }=\sqrt{2m_{\alpha }{\left(KE\right)}_{\alpha }} \\ \Rightarrow p_{\alpha }=\sqrt{2m_{\alpha }\left(q_{\alpha }V\right)} \end{gathered}$$
For proton-
$$\begin{gathered} \Rightarrow p_p=\sqrt{2m_p{\left(KE\right)}_p} \\ \Rightarrow p_p=\sqrt{2m_p\left(q_{p}V\right)} \end{gathered}$$
The formula for de-Broglie wavelength is $\lambda ={\displaystyle \frac{h}{p}}$. Putting the values of momentum from above one by one, we get-
For $\alpha$ particle-
$\Rightarrow {\lambda }_{\alpha }={\displaystyle \frac{h}{\sqrt{2m_{\alpha }\left(q_{\alpha }V\right)}}}$
For proton-
$\Rightarrow {\lambda }_p={\displaystyle \frac{h}{\sqrt{2m_p\left(q_{p}V\right)}}}$
Finding their ratios, we will have-
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{{\displaystyle \frac{h}{\sqrt{2m_{\alpha }\left(q_{\alpha }V\right)}}}}{{\displaystyle \frac{h}{\sqrt{2m_p\left(q_{p}V\right)}}}}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{h\times \sqrt{2m_p\left(q_{p}V\right)}}{h\times \sqrt{2m_{\alpha }\left(q_{\alpha }V\right)}}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{\sqrt{m_p{q}_p}}{\sqrt{m_{\alpha }{q}_{\alpha }}}} \end{gathered}$$
As we know that the mass of $\alpha$ particle is 4 times the mass of proton and the charge of $\alpha$ particle is 2 times the charge of proton, we get-
$$\begin{gathered} \Rightarrow m_{\alpha }=4m_p \\ \Rightarrow q_{\alpha }=2q_p \end{gathered}$$
Putting the values in the above ratio, we will have-
$$\begin{gathered} \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{\sqrt{m_p{q}_p}}{\sqrt{m_{\alpha }{q}_{\alpha }}}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{\sqrt{m_p{q}_p}}{\sqrt{4m_{p}2q_p}}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}=\sqrt{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{1}{2}}} \\ \Rightarrow {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_p}}={\displaystyle \frac{1}{2\sqrt{2}}} \end{gathered}$$
Thus, the ratio of ${\lambda }_{\alpha }:{\lambda }_p=1:2\sqrt{2}$.

Note: It is said that matter has a dual nature of wave-particles. de Broglie waves, named after the pioneer Louis de Broglie, is the property of a material object that differs in time or space while acting like waves. It is likewise called matter-waves. It holds extraordinary likeness to the dual nature of light which acts as particle and wave, which has been demonstrated experimentally.