Question

Calculate de Broglie wavelength of an electron travelling at $1\%$ of the speed of light

• A. $2.73\times 10^{-24}$
• B. $2.42\times 10^{-10}$
• C. $242.2\times 10^{10}$
• D. None of these

Answer 1

Answer: (B)

One percent of the speed of light is
$v=\left(\frac{1}{100}\right)\left(3.00\times 10^{8}m{s}^{-1}\right)$
$=\left(3.00\times 10^{8}m{s}^{-1}\right)$
Momentum of the electron $\left(p\right)=mv$
$=\left(9.11\times 10^{-31}Kg\right)\left(3.00\times 10^{6}m{s}^{-1}\right)$
The de-Broglie wavelength of this electron is
$\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{2.73\times 10^{-24}Kgm{s}^{-1}}$
$\lambda =2.424\times 10^{-10}m.$