Question

The de Broglie wavelength (in nanometer) for an electronmoving with 18.72 x 104 m/s is:

Answer 1

The Broglie wave equation is

$\lambda=\frac{h}{p}$

Where p is the momentum (m*v) and h = 6.63x10^-34 J*s.

Find the momentum of an electron first.

$p=9.11\times10^{-31}\operatorname{kg}\cdot18.72\times10^4\cdot\frac{m}{s}=1.71\times10^{-25}\operatorname{kg}\cdot\frac{m}{s}$

Then, find the Broglie wavelength.

$\lambda=\frac{6.63\times10^{-34}J\cdot s}{1.71\times10^{-25}\operatorname{kg}\cdot\frac{m}{s}}=3.88\times10^{-9}m$

Therefore, the Broglie wavelength is 3.88x10^-9 meters.