Question

The radio of de-Broglie wavelength of a proton and an alpha particle of same energy is
A. 1
B. 2
C. 4
D. 0.25

Answer 1

Hint:
-Matter has a dual nature of wave-particles.
-de Broglie waves, named after Louis de Broglie.
-The property of a material object varies in time or space while behaving similar to waves.
-It is also called matter-waves.

Formula used:
$\lambda = \dfrac{{\text{h}}}{{\text{p}}}$, here $\lambda $= wavelength of wave, h=Planck’s constant, p=momentum
${\text{p = mv}}$, here m=particle mass, v=velocity of the particle
${\text{p = }}\sqrt {2{\text{mK}}{\text{.E}}{\text{.}}} $, here K.E. =kinetic energy of the particle

Complete step-by-step solution:
According to de-Broglie $\lambda = \dfrac{{\text{h}}}{{\text{p}}}$
${\lambda _{\text{p}}} = $ de-Broglie wavelength of proton particle, ${\lambda _{\text{p}}} = \dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{p}}}{\text{K}}{\text{.E}}{\text{.}}} }}$
${\lambda _\alpha } = $de-Broglie wavelength of alpha particle, ${\lambda _\alpha } = \dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_\alpha }{\text{K}}{\text{.E}}{\text{.}}} }}$
K.E.= kinetic energy of the particle
Now, we will find the ratio of de-Broglie wavelength, $\dfrac{{{\lambda _\alpha }}}{{{\lambda _{\text{p}}}}} = \dfrac{{\dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_\alpha }{\text{K}}{\text{.E}}{\text{.}}} }}}}{{\dfrac{{\text{h}}}{{\sqrt {2{{\text{m}}_{\text{p}}}{\text{K}}{\text{.E}}{\text{.}}} }}}}$
We know that the mass of alpha particle is four times the mass of proton, ${{\text{m}}_\alpha } = 4{{\text{m}}_{\text{p}}}$ and
K.E. of proton=K.E. of alpha particle
Cancelling all the terms and substituting the values, we get,
$\dfrac{{{\lambda _\alpha }}}{{{\lambda _{\text{p}}}}} = \dfrac{{\sqrt {{{\text{m}}_\alpha }} }}{{\sqrt {{{\text{m}}_{\text{p}}}} }}$
$ \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _{\text{p}}}}} = \dfrac{{\sqrt {4{{\text{m}}_{\text{p}}}} }}{{\sqrt {{{\text{m}}_{\text{p}}}} }}$
$ \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _{\text{p}}}}} = \sqrt {\dfrac{4}{1}} $
$ \Rightarrow \dfrac{{{\lambda _\alpha }}}{{{\lambda _{\text{p}}}}} = \dfrac{2}{1}$

Hence the correct option is \[\left( B \right)\]

Note:
-It has been proven experimentally that the dual nature of light which behaves as particle and wave.
-The physicist Louis de Broglie suggested that particles might have wave properties and particle properties as well.
-The objects in day-to-day life have wavelengths which are very small and invisible, hence, we do not experience them as waves.
-de Broglie wavelengths are quite visible in subatomic particles.
-The electrons in atoms circle the nucleus in specific configurations, or states, which are called stationary orbits.