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The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:
A
13.6 nm
B
136 nm
C
1.36 nm
D
0.136 nm
Detailed Solution
$\lambda_e=\dfrac{12.27}{\sqrt V}$ Å $=\dfrac{12.27}{\sqrt{81}}=\dfrac{12.27}{9}=1.36$ Å $=0.136$ nm (since $1$ Å $=\dfrac{1}{10}$ nm).