Atomic Structure Question 499
Question: Calculate de-Broglie wavelength of an electron travelling at 1% of the speed of light
[DPMT 2004]
Options:
A) $ 2.73\times {10^{-24}} $
B) $ 2.42\times {10^{-10}} $
C) $ 242.2\times 10^{10} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
One percent of the speed of light is $ v=( \frac{1}{100} )(3.00\times 10^{8}m{s^{-1}}) $ = $ 3.00\times 10^{6}m{s^{-1}} $ Momentum of the electron $ (p) $ = $ m\nu $ = $ (9.11\times {10^{-31}}kg),(3.00\times 10^{6}m{s^{-1}}) $ = $ 2.73\times {10^{-24}}kg,m{s^{-1}} $ The de-broglie wavelength of this electron is $ \lambda =\frac{h}{p}=\frac{6.626\times {10^{-34}}}{2.73\times {10^{-24}}kgm{s^{-1}}} $
$ \lambda =2.424\times {10^{-10}}m $ .
BETA
