Question

A proton and an $\alpha$ particle have the same de-Broglie wavelength. Determine the ratio of
(i) their accelerating potentials
(ii) their speeds.

Answer 1

If two particles have the same de-Broglie wavelength, then they have the same momentum. Also the mass of an $\alpha$ particle is $4$ times more than that of a proton.

Complete step by step solution:
Let the de-Broglie wavelength of proton be ${\lambda _p}$ and the de-Broglie wavelength of $\alpha$ particle be ${\lambda _\alpha }$
Given that a proton and an $\alpha$ particle have the same de-Broglie wavelength. Therefore
${\lambda _p} = {\lambda _\alpha }$
$\Rightarrow \dfrac{h}{{{p_p}}} = \dfrac{h}{{{p_\alpha }}}$
Here, $h$ is the Planck ’s constant, ${p_p}$ is the momentum of the proton and ${p_\alpha }$ is the momentum of $\alpha$ particle.
$\Rightarrow {p_p} = {p_\alpha }$
So, by definition of momentum we get
$\Rightarrow {m_p}{v_p} = {m_\alpha }{v_\alpha }$
$\Rightarrow \dfrac{{{v_p}}}{{{v_\alpha }}} = \dfrac{{{m_\alpha }}}{{{m_p}}}$
We know that, $\alpha$ particles are actually $H{e^{2 + }}$ ions that have 2 protons and two neutrons
So, ${m_\alpha } = 4{m_p}$
$\therefore \dfrac{{{v_p}}}{{{v_\alpha }}} = 4$
Hence the ratio of their velocities is $4:1$.

Again,
${\lambda _p} = {\lambda _\alpha }$
$\Rightarrow \dfrac{h}{{{p_p}}} = \dfrac{h}{{{p_\alpha }}}$
Cancelling the common terms from both sides and taking the reciprocal,
$\Rightarrow {p_p} = {p_\alpha }$
Now, writing momentum in terms of energy, we get;
we know that, $\dfrac{{{p^2}}}{{2m}} = qV$
thus, $p = \sqrt {2qVm}$
$\Rightarrow \sqrt {2{q_p}{V_P}{m_p}} = \sqrt {2{q_\alpha }{V_\alpha }{m_\alpha }}$
Squaring both sides
$\Rightarrow 2{q_p}{V_p}{m_p} = 2{q_\alpha }{V_\alpha }{m_\alpha }$
Cancelling the common terms from both sides,
$\Rightarrow {q_p}{V_p}{m_p} = {q_\alpha }{V_\alpha }{m_\alpha }$
$\Rightarrow \dfrac{{{V_p}}}{{{V_\alpha }}} = \dfrac{{{q_\alpha }{m_\alpha }}}{{{q_p}{m_p}}}$
Since, ${m_\alpha } = 4{m_p}$ and ${q_\alpha } = 2{q_p}$
$\Rightarrow \dfrac{{{V_p}}}{{{V_\alpha }}} = 2 \times 4$
$\therefore \dfrac{{{V_p}}}{{{V_\alpha }}} = 8$
Hence the ratio of their accelerating potential is $8:1$.

Additional Information:
It is said that matter has a dual nature of wave-particles. De Broglie waves, named after the discoverer Louis de Broglie, is the property of a material object that varies in time or space while behaving similar to waves. It is also called matter-waves. It holds great similarity to the dual nature of light which behaves as particle and wave, which has been proven experimentally.
The physicist Louis de Broglie suggested that particles might have both wave properties and particle properties. The wave nature of electrons was also detected experimentally to substantiate the suggestion of Louis de Broglie.

The objects which we see in day-to-day life have wavelengths which are very small and invisible; hence, we do not experience them as waves. However, de Broglie wavelengths are quite visible in the case of subatomic particles.
Note: Momentum can be expressed in different ways. It can be expressed as $p = mv$ , $p = \sqrt {2Km}$ and $p = \sqrt {2qVm}$ where $m$ is the mass of the particle, $v$ is the velocity, $q$ is the charge if the particle is charged, $V$ is the accelerating potential and $K$ is the Kinetic Energy of the particle.