Question

The de Broglie wavelength of an electron traveling at 7.0 x 107m/s is 1.0 x 10-11m. (Remember that me = 9.1 x 10-31kg and h = 6.63 x 10-34J·s)

True
False

Answer 1

The De Broglie's wavelength of a particle is given by:

`\lambda=(h)/(p)`

where

`h=6.6 \cdot 10^(-34) Js` is the Planck constant

p is the momentum of the particle

In this problem, the momentum of the electron is equal to the product between its mass and its speed:

`p=m_e v=(9.1 \cdot 10^(-31) kg)(7.0 \cdot 10^7 m/s)=6.4 \cdot 10^(-23) kg m/s`

and if we substitute this into the previous equation, we find the De Broglie wavelength of the electron:

`\lambda=(h)/(p)=(6.6 \cdot 10^(-34) Js)/(6.4 \cdot 10^(-23) kg m/s)=1.0 \cdot 10^(-11) m`

So, the answer is True.