Physics - Modern Physics Question with Solution | TestHub

De-Broglie wavelength of an electron is given by

$\mathrm{\lambda }=\frac{\mathit{h}}{\mathit{m}\mathit{v}}=\frac{h}{\sqrt{2meV}}-\frac{12.27}{\sqrt{V}} \mathrm{Å}$

($h=$Planck's constant)

Where,$m=$mass of electron,$e=$electronic charge and$V=$potential difference with which electron is accelerated.

$\mathrm{\lambda }=\frac{12.27}{\sqrt{100}} \mathrm{Å}=\frac{12.27}{10} \mathrm{Å}$

$=1.227\mathrm{ }\mathrm{Å}$