Question

The wavelength of a charged particle of mass $8.0 \times 10^{-31} kg$, charge $1.6 \times 10^{- 19} C$ and kinetic energy $3 \,keV$ will be (Planck constant, $\left.h=6.4 \times 10^{-34} Js \right)$

• A. $0.4\, \mathring{A}$
• B. $2.1 \, \mathring{A}$
• C. $1.0 \, \mathring{A}$
• D. $1\, \mathring{A}$

Answer 1

Answer: (A)

Here, $m=8 \times 10^{-31} kg , q=1.6 \times 10^{-19} \,C$
and $KE =3 keV$
Now, the de-Broglie wavelength, of a charged particle,
$\lambda=\frac{h}{p}\,\,\,\,\,\dots(i)$
and $p=\sqrt{2 m KE }\,\,\,\,\,\dots(ii)$
From Eq. (i) and (ii), we get
$\lambda=\frac{h}{\sqrt{2 m KE }}$
Putting the given values, we get
$\Rightarrow \, \lambda=\frac{6.4 \times 10^{-34}}{\sqrt{2 \times 8 \times 10^{-31} \times 3 \times 10^{3} \times 1.6 \times 10^{-19}}}$
$\Rightarrow \, \lambda \approx 0.4 \, \mathring{A}$