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  4. The de-Broglie wavelength and kinetic energy of a particle is 2000 mathringA and 1 eV respectively. If its kinetic energy becomes 1 MeV, then its de Broglie wavelength

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Q. The de-Broglie wavelength and kinetic energy of a particle is $2000 \,\mathring{A}$ and $1\, eV$ respectively. If its kinetic energy becomes $1\, MeV$, then its de Broglie wavelength

KEAMKEAM 2013Dual Nature of Radiation and Matter

Solution:

Given, $\lambda=2000 \,\mathring{A} KE _{1}=1 \,eV$ and $KE _{2}=1 \,MeV$
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 } $
$\lambda =\frac{2000 \times \sqrt{1 eV }}{\sqrt{10^{6} eV }}$
$\lambda =2 \, \mathring{A}$