a sample of atoms (molar mass = 153. g/mol) is accelerated to 60.0% of the speed of light. what is the de…

Explanation:

Step1: Calculate the mass of one - mole of atoms in kg

The molar mass of the atoms is $M = 153\ \frac{g}{mol}$. Convert grams to kilograms: $M=153\times10^{- 3}\ kg/mol$. Using Avogadro's number $N_A = 6.022\times10^{23}\ mol^{-1}$, the mass of one atom $m=\frac{M}{N_A}=\frac{153\times10^{-3}\ kg/mol}{6.022\times10^{23}\ mol^{-1}}$.

Step2: Calculate the speed of the atoms

The atoms are accelerated to $v = 0.600c$, where $c = 3\times10^{8}\ m/s$. So $v=0.600\times3\times10^{8}\ m/s = 1.8\times10^{8}\ m/s$.

Step3: Use the de - Broglie wavelength formula

The de - Broglie wavelength formula is $\lambda=\frac{h}{p}$, and since $p = mv$ (for non - relativistic cases, and for speeds close to the speed of light we can still use the non - relativistic formula for a rough estimate here as the speed is not extremely close to $c$), $\lambda=\frac{h}{mv}$, where $h = 6.63\times10^{-34}\ J\cdot s$. First, calculate $m=\frac{153\times10^{-3}}{6.022\times10^{23}}\ kg\approx2.54\times10^{-25}\ kg$. Then $\lambda=\frac{6.63\times10^{-34}\ J\cdot s}{(2.54\times10^{-25}\ kg)\times(1.8\times10^{8}\ m/s)}$.

Step4: Simplify the expression for the wavelength

$\lambda=\frac{6.63\times10^{-34}}{2.54\times10^{-25}\times1.8\times10^{8}}\ m=\frac{6.63\times10^{-34}}{4.572\times10^{-17}}\ m\approx1.45\times10^{-17}\ m$. Convert to nm: $\lambda = 1.45\times10^{-8}\ nm$.

Answer:

$1.45\times10^{-8}\ nm$