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 $=6.6\times {10}^{-34}$Js) (nearly)

• A. 140 $A^0$
• B. 0.14 $A^0$
• C. 14 $A^0$
• D. 1.4$A^0$

Answer 1

Answer: (D)

1.4$A^0$

$given,\frac{1}{2}m{v}^2=80\times 1.6\times {10}^{-19}$

$mv=\sqrt{2\times 9\times {10}^{-3.1}\times 80\times 1.6\times {10}^{-19}}$

$\therefore \lambda =\frac{h}{mv}$

$=\frac{6.6\times {10}^{34}}{\sqrt{2\times 9\times {10}^{-31}\times 80\times 1.6\times {10}^{-19}}}$

$=1.4A^0$

So, the answer is option (D).