Question

An electron, a proton and an alpha particle have KE of 16E, 4E and E respectively. What is the qualitative order of their de-Broglie wavelength?
A. \[{\lambda _{\text{e}}} > {\lambda _{\text{p}}} = {\lambda _\alpha }\]
B. \[{\lambda _{\text{p}}} = {\lambda _\alpha } > {\lambda _{\text{e}}}\]
C. \[{\lambda _{\text{p}}} > {\lambda _{\text{e}}} > {\lambda _\alpha }\]
D. \[{\lambda _\alpha } = {\lambda _{\text{e}}} \gg {\lambda _{\text{p}}}\]

Answer 1

Hint: The de-Broglie equation can be used to correlate de-Broglie wavelength, momentum, mass and kinetic energy of particles of matter. Also, de-Broglie wavelength is determined for matter waves which explains wave nature of matter.
Formula Used: \[\lambda = \dfrac{{\text{h}}}{{\text{p}}} = \dfrac{{\text{h}}}{{\sqrt {2{\text{mKE}}} }}\] ; here h represents Planck’s constant, p represents momentum, m represent mass, v is velocity, KE is kinetic energy of particle of matter.

Complete step by step answer:
As light has dual nature. Light possesses both particle and wave nature. De-Broglie suggested that matter could also have both nature; particles and wave nature. As we all already know, particle nature and properties of particles of matter like they have space between them, particles of a matter attract each other; they show movement about their mean position. But de-Broglie suggested the matter behavior of matter and explained it as matter waves. So, matter waves are the imaginary waves associated with material or matter particles. The wavelength of matter waves is known as de-Broglie wavelength.
The mass of the proton is approximately 1800 times the mass of the electron. \[{{\text{m}}_{\text{p}}} = 1.67 \times {10^{ - 27}}{\text{Kg}} \approx 1800 \times {{\text{m}}_{\text{e}}}\] . The mass of the \[\alpha \] particle is twice the mass of the proton. \[{{\text{m}}_\alpha } = 4{{\text{m}}_{\text{p}}}\] .
Now as given, an electron, a proton and an alpha particle have KE of 16E, 4E and E respectively. Their de-Broglie wavelength can be given as:
De-Broglie wavelength of electron having energy 16E can be given as:
 \[{\lambda _{\text{e}}} = \dfrac{{\text{h}}}{{\sqrt {2 \times \left( {\dfrac{{{{\text{m}}_{\text{p}}}}}{{1800}}} \right) \times 16{\text{E}}} }}\]
De-Broglie wavelength of proton having energy 4E can be given as:
 \[{\lambda _{\text{p}}} = \dfrac{{\text{h}}}{{\sqrt {2 \times {{\text{m}}_{\text{p}}} \times 4{\text{E}}} }}\]
De-Broglie wavelength of \[\alpha \] particle having energy E can be given as:
 \[{\lambda _\alpha } = \dfrac{{\text{h}}}{{\sqrt {2 \times \left( {4{{\text{m}}_{\text{p}}}} \right) \times {\text{E}}} }}\]
On solving we get order of de-Broglie wavelength: \[{\lambda _{\text{e}}} > {\lambda _{\text{p}}} = {\lambda _\alpha }\] .

Therefore, the correct option is A.
Note:
An \[\alpha \] particle has twice the charge of the proton and 4 times the mass of the proton. An \[\alpha \] can be considered as a Helium nucleus because it has two protons which give it \[ + 2\] charge and also 2 neutrons which along with two protons give a mass of 4amu.