Question

A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has 1) greater value of de-Broglie wavelength, associated with it and 2) less kinetic energy? Explain.

Answer 1

Use conservation of energy which states that change in electrostatic energy equals to the kinetic energy gained and for second solution use relation which gives relation between energy, charge and potential applied. In the first part, compare the wavelength of the deuterium and alpha particle and in the second part, compare the kinetic energy of alpha particle and deuterium.
Formula used:
According to de Broglie’s hypothesis, moving electrons are associated with wavelength which is given by-
$\lambda =\dfrac{h}{p}$
Where,
h is planck's constant
p is momentum of the electron

** Complete step-by-step answer :**
If the electron of mass $\text{m}$ is accelerated by a potential difference V, the work done on the electron increases its kinetic energy. The energy of electron is given by-
$\text{E=qV}$
Also,
$\text{E=}\dfrac{1}{2}m{{v}^{2}}$
$E=\dfrac{1}{2}\dfrac{{{m}^{2}}}{m}{{v}^{2}}=\dfrac{{{\left( mv \right)}^{2}}}{2m}=\dfrac{{{p}^{2}}}{2m}$
$p=\sqrt{2mqV}$
Then $\lambda =\dfrac{h}{\sqrt{2mqV}}$

1. Let, $\text{ }{{\lambda }_{d}},{{m}_{d}}$and ${{\text{q}}_{d}}$ are wavelength, mass and charge of deuterium
   Let, ${{\lambda }_{\alpha }},{{m}_{\alpha }}$ are ${{\text{q}}_{\alpha }}$ are wavelength, mass and change of alpha particle
   We know that, wavelength of deuterium is given by,
   ${{\lambda }_{d}}=\dfrac{h}{\sqrt{2{{m}_{d}}{{q}_{d}}{{V}_{d}}}}---(1)$
   And wavelength of alpha particle is given by,
   ${{\lambda }_{\alpha }}=\dfrac{h}{\sqrt{2{{m}_{\alpha }}{{q}_{\alpha }}{{V}_{\alpha }}}}---(2)$
   Take the ratio of equation (1) and (2) and cancel h and V because h and V both are constant. Therefore. We get
   $\dfrac{{{\lambda }_{d}}}{{{\lambda }_{\alpha }}}=\dfrac{\sqrt{{{m}_{\alpha }}{{q}_{\alpha }}}}{\sqrt{{{m}_{d}}{{q}_{d}}}}$
   We know that mass and charge of deuterium are 2 and 1 respectively and mass and charge of alpha particle are 4 and 2 respectively
   Therefore, we get

& \dfrac{{{\lambda }_{d}}}{{{\lambda }_{\alpha }}}=\dfrac{\sqrt{4\times 2}}{\sqrt{2\times 1}}=2 \\\ & {{\lambda }_{d}}=2{{\lambda }_{\alpha }} \\\ \end{aligned}$$ From the above equation we can say that the wavelength of deuterium is twice than the wavelength of an alpha particle. Hence, the wavelength of a deuteron is more than the wavelength of an alpha particle. 2) As we discussed kinetic energy can also be written as $\text{E=qV}$ $\begin{aligned} & \dfrac{{{E}_{d}}}{{{E}_{\alpha }}}=\dfrac{{{q}_{d}}}{{{q}_{\alpha }}}=\dfrac{1}{2} \\\ & {{E}_{\alpha }}=2{{E}_{d}} \\\ \end{aligned}$ We can say that the kinetic energy of an alpha particle is twice the kinetic energy of deuterium. Hence the kinetic energy of an alpha particle is less than the kinetic energy of a deuteron. **Note** : We know that the mass and charge of deuterium are 2 and 1 respectively and the mass and charge of alpha particles are 4 and 2 respectively. The wavelength of any particle is inversely proportional to the mass of the particle, charge on particle and voltage applied, and Planck’s constant. So we can say that the more the value of the mass of the particle, charge on the particle, and the voltage applied, the lesser will be the value of wavelength. In this question, the mass and charge of alpha particles are more than deuterium, therefore the wavelength of an alpha particle is lesser than deuterium because of an inversely proportional relation. Similarly, for the second part, kinetic energy depends on the charge of the particle and voltage applied. They have directly proportional relations.