MCQ
The de-Broglie wavelength $\lambda $ associated with an electron having kinetic energy $E$ is given by the expression
- ✓$\frac{h}{{\sqrt {2mE} }}$
- B$\frac{{2h}}{{mE}}$
- C$2mhE$
- D$\frac{{2\sqrt {2mE} }}{h}$
✓
Answer
Correct option: A.
$\frac{h}{{\sqrt {2mE} }}$
a
(a) $\frac{1}{2}m{v^2} = E $
(a) $\frac{1}{2}m{v^2} = E $
$\Rightarrow mv = \sqrt {2mE} ;\;\;\therefore \;\;\lambda = \frac{h}{{mv}} = \frac{h}{{\sqrt {2mE} }}$
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