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 $3keV$ will be (Planck constant, $h=6.4\times 10^{-34}Js)$

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