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Question 10

An atom of helium has a radius 

rHe

 = 

31.
 
pm

 and an average speed in the gas phase at 

25

°C

 of 

787

.⁢

/ms

.

Suppose the speed of a helium atom at 

25

°C

 has been measured to within 

0.10%

. Calculate the smallest possible length of box inside of which the atom could be known to be located with certainty.

Write your answer as a multiple of 

rHe

 and round it to 

2

 significant figures. For example, if the smallest box the atom could be in turns out to be 

42.0

 times the radius of an atom of helium, you would enter "

42.

rHe

" as your answer.


To find the smallest possible length of the box inside which a helium atom could be located with certainty, we can use the uncertainty principle from quantum mechanics. The principle states:

\[
\Delta x \Delta p \geq \frac{\hbar}{2}
\]

where:
- \(\Delta x\) is the uncertainty in position,
- \(\Delta p\) is the uncertainty in momentum,
- \(\hbar\) is the reduced Planck's constant (\(\approx 1.055 \times 10^{-34} \, \text{Js}\)).

1. **Calculate the momentum (\(p\)) of a helium atom:**
   \[
   p = m \cdot v
   \]
   where \(m\) is the mass of the helium atom and \(v\) is its speed.

   The molar mass of helium is approximately \(4 \, \text{g/mol}\). The mass of one helium atom is:
   \[
   m = \frac{4 \times 10^{-3} \, \text{kg}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 6.646 \times 10^{-27} \, \text{kg}
   \]

   The speed (\(v\)) is given as \(787 \, \text{m/s}\).

   Thus, the momentum \(p\) is:
   \[
   p = (6.646 \times 10^{-27} \, \text{kg}) \cdot (787 \, \text{m/s}) \approx 5.24 \times 10^{-24} \, \text{kg m/s}
   \]

2. **Calculate the uncertainty in momentum (\(\Delta p\)):**
   The speed is measured to within \(0.10\%\):
   \[
   \Delta v = 0.001 \times 787 \, \text{m/s} \approx 0.787 \, \text{m/s}
   \]
   Therefore, the uncertainty in momentum is:
   \[
   \Delta p = m \cdot \Delta v = (6.646 \times 10^{-27} \, \text{kg}) \cdot (0.787 \, \text{m/s}) \approx 5.24 \times 10^{-27} \, \text{kg m/s}
   \]

3. **Apply the uncertainty principle:**
   Rearranging the uncertainty principle gives:
   \[
   \Delta x \geq \frac{\hbar}{2 \Delta p}
   \]
   Plugging in values:
   \[
   \Delta x \geq \frac{1.055 \times 10^{-34} \, \text{Js}}{2 \cdot 5.24 \times 10^{-27} \, \text{kg m/s}} \approx 1.01 \times 10^{-8} \, \text{m} \approx 10.1 \, \text{nm}
   \]

4. **Convert \(\Delta x\) to a multiple of \(r_{He}\):**
   The radius of a helium atom is given as \(31 \, \text{pm} = 3.1 \times 10^{-11} \, \text{m}\). Therefore, we calculate:
   \[
   \frac{\Delta x}{r_{He}} = \frac{10.1 \times 10^{-9} \, \text{m}}{3.1 \times 10^{-11} \, \text{m}} \approx 325.8
   \]

   Rounding to 2 significant figures gives:
   \[
   \Delta x \approx 330 \, r_{He}
   \]

Thus, the smallest possible length of the box is:

\[
\boxed{330 \, r_{He}}
\]

Would you like any further details or have questions about the calculations? Here are some related questions you might consider:

1. How does the uncertainty principle apply to other particles?
2. What are the implications of this uncertainty for quantum mechanics?
3. How would the calculation change for a different gas?
4. What other methods are used to measure atomic sizes?
5. Can the uncertainty principle be observed directly in experiments? 

Tip: Always ensure to keep track of units in physics problems to avoid errors in calculations!

Calculate the smallest possible length of the box inside of which a helium atom could be known to be located with certainty.

Explore how to calculate the smallest possible length of a box that can contain a helium atom with certainty, using quantum mechanics principles and the uncertainty principle. This solution involves key formulas, including the momentum of the atom and uncertainties in position and momentum.

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