Question

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

Answer 1

Hint- Here, we will proceed by writing down the de-Broglie wavelength of the proton in terms of its mass, charge and velocity. Then, we will represent the de-Broglie wavelength of the alpha particle in terms of the mass, charge and velocity of the proton. Finally, we will divide these two.

Step-By-Step answer:
Formulas Used- ${{\text{m}}_\alpha } = 4{{\text{m}}_{\text{p}}}$, ${{\text{q}}_\alpha } = 2{{\text{q}}_{\text{p}}}$ and $\lambda = \dfrac{{\text{h}}}{{\sqrt {2{\text{mqv}}} }}$.
As we know that the mass of the alpha particle ${{\text{m}}_\alpha }$ is equal to four times the mass of the proton ${{\text{m}}_{\text{p}}}$ i.e., ${{\text{m}}_\alpha } = 4{{\text{m}}_{\text{p}}}{\text{ }} \to {\text{(1)}}$
Also, the charge on an alpha particle ${{\text{q}}_\alpha }$ is equal to two times the charge on the proton ${{\text{q}}_{\text{p}}}$ i.e., ${{\text{q}}_\alpha } = 2{{\text{q}}_{\text{p}}}{\text{ }} \to {\text{(2)}}$
Since, the de-Broglie wavelength of any particle is given by
$\lambda = \dfrac{{\text{h}}}{{\sqrt {2{\text{mqv}}} }}{\text{ }} \to {\text{(3)}}$ where $\lambda $ denotes the de-Broglie wavelength of the particle, h denotes the Planck’s constant and is equal to $6.6 \times {10^{ - 34}}$ Joule-sec (Js), m denotes the mass of the particle, q denotes the charge on the particle and v denotes the velocity with which the particle moves
It is given that both the proton and the alpha particle are accelerated under the same potential difference which means that the velocity of both these particles will be equal.
i.e., Velocity of alpha particle = Velocity of proton = v (assume)
Using equation (3), the de-Broglie wavelength of the proton is given by
${\lambda _{\text{p}}} = \dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }}{\text{ }} \to {\text{(4)}}$
Using equation (3), the de-Broglie wavelength of the alpha particle is given by
${\lambda _\alpha } = \dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_\alpha }{{\text{q}}_\alpha }{\text{v}}} }}$
Using equations (1) and (2) in the above equation, we get
$
  {\lambda _\alpha } = \dfrac{{\text{h}}}{{\sqrt {2\left( {4{{\text{m}}_{\text{p}}}} \right)\left( {2{{\text{q}}_{\text{p}}}} \right){\text{v}}} }} \\
   \Rightarrow {\lambda _\alpha } = \dfrac{{\text{h}}}{{\sqrt {16{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }}{\text{ }} \to {\text{(5)}} \\
 $
By dividing equation (4) by equation (5), we get
\[
  \dfrac{{{\lambda _{\text{p}}}}}{{{\lambda _\alpha }}} = \dfrac{{\dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }}}}{{\dfrac{{\text{h}}}{{\sqrt {16{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }}}} \\
   \Rightarrow \dfrac{{{\lambda _{\text{p}}}}}{{{\lambda _\alpha }}} = \dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }} \times \dfrac{{\sqrt {16{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}} }}{{\text{h}}} \\
   \Rightarrow \dfrac{{{\lambda _{\text{p}}}}}{{{\lambda _\alpha }}} = \sqrt {\dfrac{{16{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}}}{{2{{\text{m}}_{\text{p}}}{{\text{q}}_{\text{p}}}{\text{v}}}}} \\
   \Rightarrow \dfrac{{{\lambda _{\text{p}}}}}{{{\lambda _\alpha }}} = \sqrt 8 \\
 \]
Therefore, the required ratio of de-Broglie wavelengths of the proton and the alpha particle is \[\sqrt 8 \].
Hence, option A is correct.

Note- A proton is simply referred to as a subatomic particle having positive charge on them. The mass of protons is slightly less than that of neutrons. Protons and neutrons are collectively termed as nucleons. An alpha particle consists of two protons and two neutrons (helium nucleus) which moves very fast.