Question

The de-Broglie wavelength and kinetic energy of a particle is 2000 and $1 \mathrm{eV}$ respectively. If its kinetic energy becomes $1 \mathrm{MeV}$, then its de-Broglie wavelength is
A.$2\dot{A}$
B.$1\dot{A}$
C.$4\dot{A}$
D.$10\dot{A}$
E.$5A$

Answer 1

Matter waves, being an example of wave-particle duality, are a central part of quantum mechanics theory. All matter displays wave-like conduct. For instance, just like a beam of light or a water wave, a beam of electrons can be diffracted. Calculate Energy level($E_1$) and $E_2$ by using de Broglie equation Then take the ratio of both the energy levels and then calculate the wavelength.

Formula used:
$\lambda=\dfrac{\mathrm{h}}{\mathrm{p}}$

Complete Step-by-Step solution:
As light exhibits both wave-like and particle-like properties, De Broglie suggested that matter exhibits wave-like and particle-like properties as well. De Broglie derived a relationship between wavelength and momentum of matter on the basis of his observations. The de Broglie relationship is known as this relationship.

The de-Broglie wavelength, $\lambda=\dfrac{\mathrm{h}}{\mathrm{p}}$
where $\mathrm{h}=$ Plank's constant and $\mathrm{p}=$ momentum of the particle Kinetic energy, $\mathrm{E}=\dfrac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\dfrac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$
As mass $m$ and $h$ are constants in both case, so $\mathrm{E} \propto \dfrac{1}{\lambda^{2}}$

Thus, $\dfrac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\dfrac{\lambda_{2}^{2}}{\lambda_{1}^{2}}$
or $\lambda _{2}^{2}=\left( \dfrac{{{E}_{1}}}{{{E}_{2}}} \right)\lambda _{1}^{2}=\dfrac{1\text{eV}}{1\times {{10}^{6}}\text{eV}}{{\left( 2000~{{\text{A}}^{{}^\circ }} \right)}^{2}}$
$\therefore {{\lambda }_{2}}=2~{{\text{A}}^{{}^\circ }}$

Its de-Broglie wavelength is ${{\lambda }_{2}}=2~{{\text{A}}^{{}^\circ }}$

Hence, the correct option is (a).

Note:
The electron starts from the rest (close enough so that $\dfrac{1}{2}m{{v}^{2}}$gives the kinetic energy gained where m is its mass and v is its speed. The electron will have a velocity of around v = 6 x 10 6 m/s for an electron gun with a voltage between its cathode and anode of $V=100v$. In quantum mechanics, the De Broglie wavelength is a wavelength manifested in all objects that determines the probability density of finding the object at a given point in the space of the configuration. A particle's de Broglie wavelength is inversely proportional to its momentum.