Answer to Question #300349 in Quantum Mechanics for Sudhir

Question

Calculate de broglie wavelength and velocity of an electron carrying kinetic energy 3keV

Answer 1

We find the speed of an electron, knowing its kinetic energy$W_k=\frac{m_e \cdot v^2}{2}$

$v=\sqrt{\frac{2 \cdot W_k}{m_e}}=\sqrt{\frac{2 \cdot 3 \cdot 10^3 \cdot 1.6 \cdot 10^{-19}}{9.1 \cdot 10^{-31}}}=3.248 \cdot 10^7 m/s$

the speed of an electron is much less than the speed of light

$v<< c$

therefore, finds the de Broglie wavelength by the formula

$\lambda_{db}=\frac{h}{m_e \cdot v}= \frac{6.62 \cdot 10^{-34}}{9.1 \cdot 10^{-31}\cdot 3.248 \cdot 10^7 }=2.24 \cdot 10^{-11}m$