Question

A proton and an alpha particle both are accelerated through the same potential difference. The ratio of the corresponding de-Broglie wavelength is:
$ (a){\text{ 2}}\sqrt 2 \\
  (b){\text{ }}\dfrac{1}{{2\sqrt 2 }} \\
  (c){\text{ 2}} \\
  (d){\text{ }}\sqrt 2 \\
 $

Answer 1

Hint – In this question use the formula for kinetic energy that is $K.E = \dfrac{1}{2}m.{v^2}$ and use that the momentum is the product of the mass of the particle and the velocity of the particle so $P = mv$. Write the formula for kinetic energy in terms of momentum and consider that the de-Broglie wavelength (\[\lambda \]) of a particle of mass m and moving with a velocity v is given as $\lambda = \dfrac{h}{{mv}}$. This will help approaching the problem.

Complete step-by-step solution -
As we know that the kinetic energy (K.E) of the particle is given as
$ \Rightarrow K.E = \dfrac{1}{2}m.{v^2}$................... (1)
Where, m = mass of the particle
              v = velocity of the particle.
Now as we all know that the momentum is the product of the mass of the particle and the velocity of the particle so we have,
$P = mv$................. (2)
Where, P = momentum
             m = mass of the particle
             v = velocity of the particle.
Now from equation (1) and (2) we have,
$ \Rightarrow K.E = \dfrac{1}{2}\dfrac{{{p^2}}}{m}$
$ \Rightarrow P = \sqrt {2m\left( {K.E} \right)} $..................... (3)
Now as we know that the de-Broglie wavelength (\[\lambda \]) of a particle of mass m and moving with a velocity v is given as,
$ \Rightarrow \lambda = \dfrac{h}{{mv}}$, where h = planck's constant.
Now from equation (2) we have,
$ \Rightarrow \lambda = \dfrac{h}{P}$
Now from equation (3) we have,
$ \Rightarrow \lambda = \dfrac{h}{{\sqrt {2m\left( {K.E} \right)} }}$..................... (4)
Now let the de-Broglie wavelength of the proton be${\lambda _1}$, mass ${m_1}$ and kinetic energy K.E = eV, where (e) is the charge on proton and v is the velocity of the proton by which it is moving so from the equation (4) we have,
$ \Rightarrow {\lambda _1} = \dfrac{h}{{\sqrt {2{m_1}eV} }}$.................. (5)
Now for an alpha particle let the de-Broglie wavelength of the alpha be ${\lambda _2}$, mass ${m_2} = 4{m_1}$ (i.e. mass of alpha particle is four times the mass of the proton) and kinetic energy K.E = 2eV, where (2e) is the charge on alpha and v is the velocity of the alpha same as proton by which it is moving so from the equation (4) we have,
$ \Rightarrow {\lambda _2} = \dfrac{h}{{\sqrt {2\left( {4{m_1}} \right)2eV} }}$.................... (6)
Now divide equation (5) by equation (6) we have,
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{h}{{\sqrt {2{m_1}eV} }}}}{{\dfrac{h}{{\sqrt {2\left( {4{m_1}} \right)2eV} }}}}$
Now simplify this we have,
$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\sqrt 8 }}{1} = 2\sqrt 2 $
So this is the required ratio of the corresponding de-Broglie wavelengths.
So this is the required answer.
Hence option (A) is the correct answer.

Note – In general an alpha particle is composed of two protons, two neutrons and is somewhat identical to a helium-4 atom. It can be produced by alpha decomposition or alpha decay of radioactive compounds. A proton is a subatomic particle. The trick point here was that the mass of the alpha particle is four times the mass of the proton.