Question

A proton and an alpha particle are accelerated under the same potential difference. Find the ratio of de-Broglie wavelengths of the proton and the alpha particle.
(A) $ \sqrt 8 $
(B) $ \sqrt {\dfrac{1}{8}} $
(C) $ 1 $
(D) $ 2 $

Answer 1

Hint: We can find the de Broglie wavelength of a proton and an alpha particle separately using the formula, $ \lambda = \dfrac{h}{p} $. Then on taking the ratio of the two values we can find the required answer.

Formula Used: The following formulas are used to solve this question.
De Broglie wavelength of a wavelength, $ {\lambda _B} = \dfrac{h}{p} = \dfrac{h}{{mv}} $ where $ h $ is Planck’s constant $ \left( {h = 6.626 \times {{10}^{ - 34}}J \cdot s} \right) $, $ p $ is the linear momentum and $ {\lambda _B} $ is the de Broglie wavelength of a particle.
 $\Rightarrow p = mv $ where $ m $ is mass and velocity of the particle is $ v $.
Kinetic energy $ K = \dfrac{1}{2}m{v^2} $
 $\Rightarrow p = \sqrt {2mK} $ where $ p $ is the linear momentum, $ m $ is mass and $ K $ is kinetic energy.
 $\Rightarrow p = \sqrt {2mqV} $ where $ q $ is charged and $ V $ is the potential difference.

Complete step by step answer
The de Broglie wavelength of an object is its wavelength $ \left( \lambda \right) $ in relation to its momentum and mass. A particle’s de Broglie wavelength is usually inversely proportional to its force.
 $ \therefore $ de Broglie wavelength of a wavelength, $ {\lambda _B} = \dfrac{h}{p} = \dfrac{h}{{mv}} $ where $ h $ is Planck’s constant $ \left( {h = 6.626 \times {{10}^{ - 34}}J \cdot s} \right) $, $ p $ is the linear momentum and $ {\lambda _B} $ is the de Broglie wavelength of a particle.
Given in the question, a proton and an alpha particle are accelerated under the same potential difference.
The mass of an $ \alpha - particle $ is $ 4 $ times the mass of a photon.
Let the mass of a photon be $ {m_p} $. Let mass of an $ \alpha - particle $ be $ {m_\alpha } = 4{m_p} $
The charge on a $ \alpha - particle $ $ \left( {{q_\alpha }} \right) $ is twice the charge on the photon particle $ \left( {{q_p}} \right) $.
 $ \therefore {q_\alpha } = 2{q_p} $
From our prior knowledge, we know that, kinetic energy $ K = \dfrac{1}{2}m{v^2} $
To relate linear momentum and kinetic energy,
 $ \therefore p = mv = \sqrt {2m \times \dfrac{1}{2}m{v^2}} $
 $ \Rightarrow p = \sqrt {2mK} $
Now, $ K = qV $.
 $ \therefore p = \sqrt {2mqV} $
The linear momentum of a $ \alpha - particle $ $ {p_\alpha } = \sqrt {2{m_\alpha }{q_\alpha }V} $
The linear momentum of a photon $ {p_p} = \sqrt {2{m_p}{q_p}V} $
 $ {\lambda _p} = \dfrac{h}{{\sqrt {2{m_p}{q_p}V} }} $
 $ {\lambda _\alpha } = \dfrac{h}{{\sqrt {2{m_\alpha }{q_\alpha }V} }} = \dfrac{h}{{\sqrt {2\left( {2{m_p}} \right)\left( {4{q_p}} \right)V} }} $
Thus the de Broglie wavelength of the $ \alpha - particle $ is given by $ {\lambda _\alpha } = \dfrac{h}{{\sqrt {16{m_p}{q_p}V} }} $
The ratio of the de-Broglie wavelengths of the proton and the alpha particle is given by $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} $.
 $ \therefore \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{{\dfrac{h}{{\sqrt {2{m_p}{q_p}V} }}}}{{\dfrac{h}{{\sqrt {16{m_p}{q_p}V} }}}} $
Cancelling the like terms, we get
 $ \Rightarrow \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \dfrac{{\sqrt {16} }}{{\sqrt 2 }} $
Thus ratio of the de-Broglie wavelengths of the proton and the alpha particle is $ \dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = \sqrt 8 $.

$ \therefore $ The correct answer is Option A.

Note
The property of a material behaving similar to waves that varies in time or space is known as matter waves. Let us consider atomic particles, like electrons that go around in circles around the atomic nuclei. In this case, the de Broglie waves exist as a closed-loop, and fit evenly around the loop. Because of this requirement, the electrons in atoms circle the nucleus in particular configurations, or states, which are called stationary orbits.
For a photon having energy $ E $ and momentum $ p $,
 $\Rightarrow E = h\nu $ where $ h $ is Planck’s constant $ \left( {h = 6.626 \times {{10}^{ - 34}}J \cdot s} \right) $ and $ c = \lambda \nu $ where $ c $ is the velocity of light in vacuum, $ \lambda $ is the wavelength and $ \nu $ is the frequency of light.
 $\Rightarrow E = \dfrac{{hc}}{\lambda } $ where $ h $ is Planck’s constant $ \left( {h = 6.626 \times {{10}^{ - 34}}J \cdot s} \right) $, $ c $ is the velocity of light in vacuum, $ \lambda $ is the wavelength and $ \nu $ is the frequency of light.