Question

missed this? watch kcv: the wave nature of matter; read section 2.4. you can click on the review link t
the smallest atoms can themselves exhibit quantum mechanical behavior.

part a

calculate the de broglie wavelength of a hydrogen atom traveling at 485 m/s.
express your answer in picometers to three significant figures.
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λ =
pm

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Explanation:

Step1: Recall de Broglie wavelength formula

The de Broglie wavelength formula is $\lambda = \frac{h}{mv}$, where $h = 6.626\times 10^{-34}\ \text{J·s}$ (Planck's constant), $m$ is the mass of the hydrogen atom, and $v$ is the velocity. The mass of a hydrogen atom (proton + electron, but electron mass is negligible here) is approximately $m = 1.674\times 10^{-27}\ \text{kg}$. The velocity $v = 485\ \text{m/s}$.

Step2: Substitute values into the formula

First, calculate the numerator and denominator separately.
$h = 6.626\times 10^{-34}\ \text{J·s}$, $m = 1.674\times 10^{-27}\ \text{kg}$, $v = 485\ \text{m/s}$.
So, $mv = 1.674\times 10^{-27}\ \text{kg} \times 485\ \text{m/s} = 1.674\times 485\times 10^{-27}\ \text{kg·m/s} \approx 811.89\times 10^{-27}\ \text{kg·m/s} = 8.1189\times 10^{-25}\ \text{kg·m/s}$.
Then, $\lambda = \frac{6.626\times 10^{-34}\ \text{J·s}}{8.1189\times 10^{-25}\ \text{kg·m/s}} \approx \frac{6.626}{8.1189}\times 10^{-9}\ \text{m} \approx 0.816\times 10^{-9}\ \text{m}$.

Step3: Convert meters to picometers

Since $1\ \text{m} = 10^{12}\ \text{pm}$, so $0.816\times 10^{-9}\ \text{m} = 0.816\times 10^{-9} \times 10^{12}\ \text{pm} = 0.816\times 10^{3}\ \text{pm} = 816\ \text{pm}$ (rounded to three significant figures).

Answer:

816