a sample of atoms (molar mass = 154. \frac{kg}{mol}) is accelerated to 60.0% of the speed of light. what is…

a sample of atoms (molar mass = 154. \frac{kg}{mol}) is accelerated to 60.0% 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 = 154. \frac{kg}{mol}) is accelerated to 60.0% 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 the mass of a single atom

First, use Avogadro's number $N_A = 6.022\times10^{23}\text{ mol}^{-1}$. The molar - mass $M = 154\text{ g/mol}=0.154\text{ kg/mol}$. The mass of a single atom $m=\frac{M}{N_A}=\frac{0.154\text{ kg/mol}}{6.022\times10^{23}\text{ mol}^{-1}}\approx2.56\times 10^{-25}\text{ kg}$.

Step2: Calculate the velocity of the atom

The speed of light $c = 3.0\times10^{8}\text{ m/s}$. The velocity of the atom $v = 0.600c=0.600\times3.0\times10^{8}\text{ m/s}=1.8\times10^{8}\text{ m/s}$.

Step3: Calculate the de - Broglie wavelength

The de - Broglie wavelength formula is $\lambda=\frac{h}{mv}$, where Planck's constant $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$. Substitute $m = 2.56\times 10^{-25}\text{ kg}$ and $v = 1.8\times10^{8}\text{ m/s}$ into the formula: $\lambda=\frac{6.626\times10^{-34}\text{ J}\cdot\text{s}}{(2.56\times 10^{-25}\text{ kg})\times(1.8\times10^{8}\text{ m/s})}\approx1.44\times10^{-17}\text{ m}$. Convert to nanometers: $\lambda = 1.44\times10^{-8}\text{ nm}$.

Answer:

$1.44\times 10^{-8}$ nm