The de-Broglie wavelength and kinetic energy of a particle is $2000 \,?$ and $1\, eV$ respectively. If its kinetic energy becomes $1\, MeV$, then its de Broglie wavelength
- $2 \,?$
- $1 \,?$
- $4 \,?$
- $10\,?$
The Correct Option is A
Solution and Explanation
We know that,
$\lambda \propto \frac{1}{\sqrt{ KE }}$
$\lambda \times \sqrt{ KE }=\text { constant }$
Hence,
$2000 \times \sqrt{1 \,eV } =\lambda \times \sqrt{10^{6} \,eV } $
$\lambda =\frac{2000 \times \sqrt{1 eV }}{\sqrt{10^{6} eV }}$
$\lambda =2 \, ?$
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Concepts Used:
De Broglie Hypothesis
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
De Broglie Equation Derivation and de Broglie Wavelength
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity ‘v’ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.