Question

What is the de Broglie wavelength of an electron that strikes the back of the face of a TV screen at 1/10 the speed of light? _____m

Answer 1

Answer: $2.42(10)^{-11} m$

Explanation:

The de Broglie wavelength $\lambda$ is given by the following formula:

$\lambda=\frac{h}{p}$   (1)

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$p$ is the momentum of the atom, which is given by:

$p=m_{e}v_{e}$   (2)

Where:

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

$v=\frac{1}{10}c$ is the velocity of the electron (we are told it is 1/10 the speed of light $c=3(10)^{8}\frac{m}{s}$)

This means equation (2) can be written as:

$p=m_{e}\frac{1}{10}c$   (3)

Substituting (3) in (1):

$\lambda=\frac{h}{m_{e}\frac{1}{10}c}$    (4)

$\lambda=10\frac{h}{m_{e} c}$    (5)

Now, we only have to find $\lambda$:

$\lambda=10\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(3(10)^{8}\frac{m}{s})}$     (6)

Finally:

$\lambda=2.42(10)^{-11} m$>>> This is the de Broglie wavelength of an electron.