Question

The de-Broglie wavelength of an electron having $80$eV of energy is nearly $(1eV=1.6\times {10}^{-19}J,$ mass of electron $=9\times {10}^{-31}kg$, Planck's constant $=6.6\times {10}^{-34}J-s)$

• A. $140\mathring{A}$
• B. $0.14\mathring{A}$
• C. $14\mathring{A}$
• D. $1.4\mathring{A}$

Answer 1

The correct option is D $1.4\mathring{A}$

$\lambda =\frac{h}{\sqrt{2mE}}$
$=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 80\times 1.6\times {10}^{-19}}}$
$=1.4\stackrel{o}{A}$