Question

During an experiment an $\alpha -$ particle and a proton are accelerated by the same potential difference, their de Broglie wavelength ratio will: (Take mass of proton $=$ mass of neutrons.)
A) $1:2$
B) $1:4$
C) $1:2\sqrt 2$
D) $1:\sqrt 2$

Answer 1

An $\alpha -$ particle is a doubly charged helium atom that consists of two protons and two neutrons bound together. Its mass is four times that of a proton and its charge is twice that of a singular proton. To solve this question, we will first find the de Broglie wavelength of both the particles and divide them to find the ratio.

Formula used:
$\lambda = \dfrac{h}{p} = \dfrac{h}{{\sqrt {2mK} }}$
Where $\lambda$ is the de Broglie wavelength, $h$ is Planck’s constant, $p$ is momentum of the particle, $m$ is the mass of the particle and $K$ is the kinetic energy.
When accelerated through the same potential difference $V$ with charge $e$ for a proton,
The kinetic energy is
$K = eV$

Complete step-by step solution:
First we begin by noting down individual masses and charges.
For a proton, the mass, ${m_p} = m$ and the charge ${q_e} = e$
Since an $\alpha -$ particle is a doubly charged helium atom that consists of two protons and two neutrons bound together, its mass is four times that of a proton and its charge is twice that of a singular proton, that is,
${m_\alpha } = 4m$
${q_\alpha } = 2e$
Therefore,
The de Broglie wavelength of a proton is
$\lambda = \dfrac{h}{p} = \dfrac{h}{{\sqrt {2mK} }}$
Where $\lambda$ is the de Broglie wavelength, $h$ is Planck’s constant, $p$ is momentum of the particle, $m$ is the mass of the particle and $K$ is the kinetic energy
When accelerated through the same potential difference $V$ with charge $e$ for a proton,
The kinetic energy is
$K = eV$
$\Rightarrow \lambda = \dfrac{h}{{\sqrt {2meV} }}$ $...(1)$
For and $\alpha -$ particle, the de Broglie wavelength is
${\lambda _\alpha } = \dfrac{h}{{\sqrt {2{m_\alpha }{q_\alpha }V} }}$
$\Rightarrow {\lambda _\alpha } = \dfrac{h}{{\sqrt {2(4m)(2e)V} }}$
$\Rightarrow {\lambda _\alpha } = \dfrac{h}{{\sqrt 8 \sqrt {2(m)(e)V} }}$ $...(2)$
Dividing $(2)$ by $(1)$
$\Rightarrow {\lambda _\alpha } = \dfrac{{\dfrac{h}{{\sqrt 8 \sqrt {2(m)(e)V} }}}}{{\dfrac{h}{{\sqrt {2meV} }}}}$
$\Rightarrow {\lambda _\alpha } = \dfrac{1}{{\sqrt 8 }} = \dfrac{1}{{2\sqrt 2 }}$

Therefore the ratio of their de Broglie wavelength is ©, $\dfrac{{{\lambda _\alpha }}}{\lambda } = 1:2\sqrt 2$.

Note: The De Broglie wavelength of a particle depends on its momentum and so depends on mass. The mass of an alpha particle is greater than that of a single proton. The wavelength is inversely proportional to the square root of mass of the particle. An Alpha particle is actually a helium ion with +2e charge. This means it only has two protons and two neutrons in this state. The mass of a neutron is approximately equal to that of a proton and so the mass of an alpha particle is 4 times the mass of a single proton.