Question
An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other?
✓
Answer
$\text{K.E.} = \frac{p^2}{2m}$$\therefore \text{K.E.} \propto \frac{1}{m} $ ( For same P)
(as $\lambda = h/ p $ is the same)
Alternate Answer
$\therefore \frac{E_ke}{E_ka} = \frac{m_a}{m}_e$
(as $\lambda = h/ p $ is the same)
Alternate Answer
$\therefore \frac{E_ke}{E_ka} = \frac{m_a}{m}_e$
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