The de Broglie wavelength associated with an electron, accelerated by a potential difference of $81…
The de Broglie wavelength associated with an electron, accelerated by a potential difference of $81 \mathrm{~V}$ is given by:
$1.36 \mathrm{~nm}$
$0.136 \mathrm{~nm}$
$13.6 \mathrm{~nm}$
$136 \mathrm{~nm}$
Solution
We know, for an electron
De Broglie wavelength, $\lambda=\frac{12.27}{\sqrt{v}} Å$
$\begin{aligned} & \lambda=\frac{12.27}{\sqrt{81}}=\frac{12.27}{9} Å \\ & \lambda=1.36 Å \text { or } 0.136 \mathrm{~nm}\end{aligned}$