Question

The de-Broglie wavelength of an electron accelerated through a potential difference of $100V$ is

• A. $12.27Å$
• B. $1.227Å$
• C. $0.1227Å$
• D. $0.001227Å$

Answer 1

Answer: (B)

De-Broglie wavelength of an electron is given by
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2meV}}-\frac{12.27}{\sqrt{V}}Å$
( $h=$ Planck's constant)
Where, $m=$ mass of electron, $e=$ electronic charge and $V=$ potential difference with which electron is accelerated.
$\lambda =\frac{12.27}{\sqrt{100}}Å=\frac{12.27}{10}Å$
$=1.227Å$