Atomic Structure Question 492
Question: An electron has kinetic energy $ 2.8\times {10^{-23}}J $ . de-Broglie wavelength will be nearly $ (m_{e}=9.1\times {10^{-31}}kg) $
Options:
A) $ 9.28\times {10^{-4}},m $
B) $ 9.28\times {10^{-7}},m $
C) $ 9.28\times {10^{-8}},m $
D) $ 9.28\times {10^{-10}},m $
Show Answer
Answer:
Correct Answer: C
Solution:
Formula for de-Broglie wavelength is $ \lambda =\frac{h}{p} $ or $ \lambda =\frac{h}{mv}\Rightarrow eV=\frac{1}{2}mv^{2} $ or $ \nu =\sqrt{\frac{2eV}{m}} $
$ \lambda =\frac{h}{\sqrt{2meV}} $
$ =\frac{6.62\times {10^{-34}}}{\sqrt{2\times 9.1\times {10^{-31}}\times 2.8\times {10^{-23}}}} $
$ \lambda =9.28\times {10^{-8}}meter $ .
BETA
