Question

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

• A. $2 \,?$
• B. $1 \,?$
• C. $4 \,?$
• D. $10\,?$

Answer 1

Answer: (A)

$2 \,?$

Answer 2

Given, $\lambda=2000 \,?, KE _{1}=1 \,eV$ and $KE _{2}=1 \,MeV$
We know that,
$\lambda \propto \frac{1}{\sqrt{ KE }}$
$\lambda \times \sqrt{ KE }=\text { constant }$
Hence,
$2000 \times \sqrt{1 \,eV } =\lambda \times \sqrt{10^{6} \,eV }$
$\lambda =\frac{2000 \times \sqrt{1 eV }}{\sqrt{10^{6} eV }}$
$\lambda =2 \, ?$