Question

The kinetic energy of an electron is $2\times {10}^{-25}J$. Calculate its de-Broglie wavelegth in nm.

$({9.1}^{1/2}\approx 3)$

Answer 1

De-Broglie wavelength $\left(\lambda \right)$ $=h/p=\frac{h}{(2mK.E{)}^{-1/2}}$

Substituting the values $\lambda =\frac{6.626\times {10}^{-34}}{(2\times 9.1\times {10}^{-31}\times 2\times {10}^{-25}{)}^{-1/2}}$

$=\frac{6.626\times {10}^{-34}}{(36\times {10}^{-6}{)}^{-1/2}}$

$=1.1\times {10}^{-31}$ m