Dual Nature Of Light Question 417
Question: The de-Broglie wavelength $ \lambda $ associated with an electron having kinetic energy E is given by the expression
Options:
A) $ \frac{h}{\sqrt{2mE}} $
B) $ \frac{2h}{mE} $
C) $ 2mhE $
D) $ \frac{2\sqrt{2mE}}{h} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{1}{2}mv^{2}=E\Rightarrow mv=\sqrt{2mE};\ \ \therefore \ \ \lambda =\frac{h}{mv}=\frac{h}{\sqrt{2mE}} $
BETA
