Answer
The wavelength of the lithium atom is $\lambda=2.279\times10^{-13}m$.
Work Step by Step
*The de Broglie relationship: $$\lambda=\frac{h}{mv}$$
$\lambda$: wavelength of object
$h$: Planck's constant ($h=6.626\times10^{-34} J.s$)
$m$: mass of object
$v$: velocity of object
1) Find the known variables
- Mass of the lithium atom:
The mass of a lithium atom is $7amu$.
$1amu=\frac{1}{6.022\times10^{23}}g\approx1.661\times10^{-24}g\approx1.661\times10^{-27}kg$
Therefore,
$7amu=7\times(1.661\times10^{-27}kg)\approx1.163\times10^{-26}kg$
In other words, $m=1.163\times10^{-26}kg$
- Velocity of the lithium atom: $v=2.5\times10^5m/s$
2) Calculate the wavelength of the lithium atom
$\lambda=\frac{h}{mv}=\frac{6.626\times10^{-34}}{(1.163\times10^{-26})\times(2.5\times10^5)}\approx2.279\times10^{-13}m$