Question

The de Broglie wavelength of an electron having $320\text{eV}$ of energy is nearly $(1\text{eV}=1.6\times {10}^{-19}\text{J}$, mass of electron $=9.1\times {10}^{-31}\text{kg}$, Planck's constant $=6.6\times {10}^{-34}\text{Js})$

• A. $70\stackrel{\circ }{\text{A}}$
• B. $0.7\stackrel{\circ }{\text{A}}$
• C. $1.4\stackrel{\circ }{\text{A}}$
• D. $14\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is B $0.7\stackrel{\circ }{\text{A}}$

de-Broglie wavelength is given by
$\lambda =\frac{h}{P}$
As $P=\sqrt{2mE}$
Hence,
$\lambda =\frac{h}{\sqrt{2mE}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 320\times 1.6\times {10}^{-19}}}$
$\lambda \approx 0.7\stackrel{\circ }{\text{A}}$