Practicing Success
Practicing Success
CUET Preparation Today
The Kinetic Energy of an electron having de Broglie wavelength($\lambda$) triple , then the de Broglie wavelength associated with it becomes: |
\(\frac{λ}{3}\) \(λ\sqrt{3}\) \(\frac{λ}{\sqrt{3}}\) 3λ |
\(\frac{λ}{\sqrt{3}}\) |
$\lambda = \frac{h}{\sqrt {2mE}}$ $\Rightarrow \lambda \propto \frac{1}{\sqrt E}$ $\lambda'= \frac{\lambda}{\sqrt 3}$ |
