Question

The wavelength associated with an electron accelerated through a potential difference of $100\, V$ is of the order of:

• A. $1.2\,\mathring{A}$
• B. $10.5\,\mathring{A}$
• C. $100\,\mathring{A}$
• D. $1000\,\mathring{A}$

Answer 1

Answer: (A)

Here : Potentiel difference $V=100\, V$
We know that de Broglie wavelength of an electron is given by as
$\lambda =\frac{h}{\sqrt{2 q V m}}$
$=\frac{6.6 \times 10^{-34}}{2 \times\left(1.6 \times 10^{-19}\right) \times 100 \times 9.1 \times 10^{-31}}$
$=1.2 \times 10^{-10} m =1.2\,\mathring{A}$