Question

Kinetic energy of electron, proton and a particle is given as K, 2K and 4K respectively, then which of the following gives the correct order of De-Broglie wavelengths of electron, proton and a particle

• A. λp > λα > λe
• B. λα > λp > λe
• C. λe > λp > λα
• D. λe > λα > λp

Answer 1

Answer: (C)

λe > λp > λα

Answer 2

$\lambda _{e}=\frac{h}{\sqrt{\frac{2\times m}{2000}}\times4K}=\frac{10\sqrt{20h}}{2\sqrt{2mK}}$
$\lambda _{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mK}}$
$\lambda _{\alpha }=\frac{h}{\sqrt{2\times 4m\times 2K}}=\frac{h}{4\sqrt{mK}}$
$\lambda _{e}>\lambda _{p}>\lambda _{\alpha }$

So, the correct option is (C): λe > λp > λα