Em Waves Question 78
Question: What is the de Broglie wavelength of the wave associated with an electron that has been accelerated through a potential difference of 50.0 V?
Options:
A) $ 2.7\times {{10}^{-10}} $
B) $ 1.74\times {{10}^{-10}} $
C) $ 3.6\times {{10}^{-9}} $
D) $ 4.9\times {{10}^{-11}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The gain of kinetic energy by an electron is eV. $ \frac{1}{2}mv^{2}=eV $
$ v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2(1.60\times {{10}^{-19}})(50)}{(9.11\times {{10}^{-31}})}} $
$ =4.19\times 10^{6}m{{s}^{-1}} $ Thus, the electron’s de Broglie wavelength is $ \lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{(9.11\times {{10}^{-31}})(4.19\times 19^{6})} $
$ =1.74\times {{10}^{-10}}m $
BETA
