a sample of atoms (molar mass = 151 $\frac{g}{mol}$) is accelerated to 45.0% of the speed of light. what is…

a sample of atoms (molar mass = 151 $\frac{g}{mol}$) is accelerated to 45.0% of the speed of light. what is the de broglie wavelength of these atoms? be sure your answer has the correct number of significant figures. note: reference the fundamental constants table for additional information. note: 1 j = $\frac{1 kg\times m^{2}}{s^{2}}$

a sample of atoms (molar mass = 151 $\frac{g}{mol}$) is accelerated to 45.0% of the speed of light. what is the de broglie wavelength of these atoms? be sure your answer has the correct number of significant figures. note: reference the fundamental constants table for additional information. note: 1 j = $\frac{1 kg\times m^{2}}{s^{2}}$

Answer

Explanation:

Step1: Calculate mass of one atom

First, convert molar - mass to mass of one atom. The molar mass $M = 151\ g/mol=0.151\ kg/mol$. Using Avogadro's number $N_A = 6.022\times10^{23}\ mol^{- 1}$, the mass of one atom $m=\frac{M}{N_A}=\frac{0.151\ kg/mol}{6.022\times10^{23}\ mol^{-1}}$. $m=\frac{0.151}{6.022\times10^{23}}\ kg\approx2.51\times10^{-25}\ kg$

Step2: Calculate the speed of the atom

The speed of light $c = 3\times10^{8}\ m/s$. The speed of the atom $v = 0.450c=0.450\times3\times10^{8}\ m/s = 1.35\times10^{8}\ m/s$.

Step3: Calculate the de - Broglie wavelength

The de - Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where Planck's constant $h = 6.626\times10^{-34}\ J\cdot s$. Substitute $m = 2.51\times10^{-25}\ kg$ and $v = 1.35\times10^{8}\ m/s$ into the formula. $\lambda=\frac{6.626\times10^{-34}\ J\cdot s}{(2.51\times10^{-25}\ kg)\times(1.35\times10^{8}\ m/s)}$ $\lambda=\frac{6.626\times10^{-34}}{2.51\times1.35\times10^{-25 + 8}}\ m$ $\lambda=\frac{6.626\times10^{-34}}{3.3885\times10^{-17}}\ m\approx1.96\times10^{-17}\ m$ Convert to nanometers: $1\ nm = 10^{-9}\ m$, so $\lambda=1.96\times10^{-8}\ nm$.

Answer:

$1.96\times10^{-8}$