Question: The de Broglie wavelength associated with an electron, accelerated by a potential difference of $81 \mathrm{~V}$ is given by:
A) $13.6 \mathrm{~nm}$
B) $136 \mathrm{~nm}$
C) $1.36 \mathrm{~nm}$
D) $0.136 \mathrm{~nm}$
Answer: $0.136 \mathrm{~nm}$
Solution:
$$ \begin{aligned} & \lambda_e=\frac{12.27}{\sqrt{V}} \stackrel{o}{A}=\frac{12.27}{\sqrt{81}} \stackrel{o}{A}=\frac{12.27}{9} \stackrel{o}{A} \ & =1.36 \stackrel{o}{A}\left(\because 1 \stackrel{o}{A}=\frac{1}{10} \mathrm{~nm}\right) \ & =0.136 \mathrm{~nm} \end{aligned} $$
