Question

The de-Broglie wavelength $\lambda$associated with an electron having kinetic energy E is given by the expression.

• A. $\frac{h}{\sqrt{2mE}}$
• B. $\frac{2h}{mE}$
• C. $2mhE$
• D. $\frac{2\sqrt{2mE}}{h}$

Answer 1

Answer: (A)

$\frac{h}{\sqrt{2mE}}$

Answer 2

$\frac{1}{2}mv^{2} = E \Rightarrow mv = \sqrt{2mE};\mspace{6mu}\mspace{6mu}\therefore\mspace{6mu}\mspace{6mu}\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mE}}$