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Q.
Calculate de Broglie wavelength of an electron travelling at $1\%$ of the speed of light
Structure of Atom
Solution:
One percent of the speed of light is
$v=\left(\frac{1}{100}\right)\left(3.00\times10^{8} ms^{-1}\right) $
$=\left(3.00\times10^{8} ms^{-1}\right)$
Momentum of the electron $\left(p\right) =\, mv$
$=\left(9.11 \times10^{-31} Kg\right)\left(3.00\times10^{6 }ms^{-1}\right) $
The de-Broglie wavelength of this electron is
$\lambda=\frac{h}{p}=\frac{6.626\times10^{-34}}{2.73\times10^{-24} Kgms^{-1}} $
$\lambda=2.424 \times10^{-10} m.$