Answer :
To solve this problem, we'll use Heisenberg's uncertainty principle, which relates the uncertainties in position and momentum. The principle can be formulated as:
[tex]\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \][/tex]
where [tex]\( \Delta x \)[/tex] is the uncertainty in position, [tex]\( \Delta p \)[/tex] is the uncertainty in momentum, and [tex]\( h \)[/tex] is Planck's constant.
1. Calculate the uncertainty in speed ([tex]\( \Delta v \)[/tex]):
Given:
- Average speed ([tex]\( v \)[/tex]) = [tex]\( 249 \, \text{m/s} \)[/tex]
- Speed uncertainty percentage = [tex]\( 0.10\% \)[/tex]
[tex]\[ \Delta v = 249 \, \text{m/s} \times \frac{0.10}{100} = 0.249 \, \text{m/s} \][/tex]
2. Calculate the uncertainty in momentum ([tex]\( \Delta p \)[/tex]):
Momentum ([tex]\( p \)[/tex]) is given by [tex]\( p = m \cdot v \)[/tex], where [tex]\( m \)[/tex] is the mass.
Let's denote the mass of an argon atom as [tex]\( m_{Ar} = 6.634 \times 10^{-26} \, \text{kg} \)[/tex].
[tex]\[ \Delta p = m_{Ar} \cdot \Delta v = 6.634 \times 10^{-26} \, \text{kg} \times 0.249 \, \text{m/s} = 1.651 \times 10^{-26} \, \text{kg m/s} \][/tex]
3. Apply Heisenberg’s uncertainty principle to find [tex]\( \Delta x \)[/tex]:
Using the reduced Planck's constant [tex]\( \hbar = \frac{h}{2\pi} \)[/tex]:
[tex]\[ \hbar = 1.0545718 \times 10^{-34} \, \text{J s} \][/tex]
From the uncertainty principle ([tex]\( \Delta x \cdot \Delta p \geq \hbar / 2 \)[/tex]):
[tex]\[ \Delta x \geq \frac{\hbar}{\Delta p} = \frac{1.0545718 \times 10^{-34} \, \text{J s}}{1.651 \times 10^{-26} \, \text{kg m/s}} \][/tex]
[tex]\[ \Delta x \approx 6.39 \times 10^{-9} \, \text{m} \][/tex]
4. Convert [tex]\( \Delta x \)[/tex] to picometers ([tex]\( \text{pm} \)[/tex]) since [tex]\( 1 \text{m} = 10^{12} \text{pm} \)[/tex]):
[tex]\[ \Delta x_{pm} = 6.39 \times 10^{-9} \, \text{m} \times 10^{12} \, \text{pm/m} = 6390 \, \text{pm} \][/tex]
5. Express the box length as a multiple of the argon atom radius ([tex]\( r_{Ar} = 71 \, \text{pm} \)[/tex]):
[tex]\[ \text{Box length in terms of } r_{Ar} = \frac{\Delta x_{pm}}{71 \, \text{pm}} \approx 89.9 \][/tex]
Rounding to 2 significant figures:
[tex]\[ \text{Smallest possible length of the box} \approx 14.31 \cdot r_{Ar} \][/tex]
Therefore, the smallest possible length of the box inside which the atom could be known to be located with certainty, expressed as a multiple of [tex]\( r_{Ar} \)[/tex], is:
[tex]\[ 14.31 \cdot r_{Ar} \][/tex]
[tex]\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \][/tex]
where [tex]\( \Delta x \)[/tex] is the uncertainty in position, [tex]\( \Delta p \)[/tex] is the uncertainty in momentum, and [tex]\( h \)[/tex] is Planck's constant.
1. Calculate the uncertainty in speed ([tex]\( \Delta v \)[/tex]):
Given:
- Average speed ([tex]\( v \)[/tex]) = [tex]\( 249 \, \text{m/s} \)[/tex]
- Speed uncertainty percentage = [tex]\( 0.10\% \)[/tex]
[tex]\[ \Delta v = 249 \, \text{m/s} \times \frac{0.10}{100} = 0.249 \, \text{m/s} \][/tex]
2. Calculate the uncertainty in momentum ([tex]\( \Delta p \)[/tex]):
Momentum ([tex]\( p \)[/tex]) is given by [tex]\( p = m \cdot v \)[/tex], where [tex]\( m \)[/tex] is the mass.
Let's denote the mass of an argon atom as [tex]\( m_{Ar} = 6.634 \times 10^{-26} \, \text{kg} \)[/tex].
[tex]\[ \Delta p = m_{Ar} \cdot \Delta v = 6.634 \times 10^{-26} \, \text{kg} \times 0.249 \, \text{m/s} = 1.651 \times 10^{-26} \, \text{kg m/s} \][/tex]
3. Apply Heisenberg’s uncertainty principle to find [tex]\( \Delta x \)[/tex]:
Using the reduced Planck's constant [tex]\( \hbar = \frac{h}{2\pi} \)[/tex]:
[tex]\[ \hbar = 1.0545718 \times 10^{-34} \, \text{J s} \][/tex]
From the uncertainty principle ([tex]\( \Delta x \cdot \Delta p \geq \hbar / 2 \)[/tex]):
[tex]\[ \Delta x \geq \frac{\hbar}{\Delta p} = \frac{1.0545718 \times 10^{-34} \, \text{J s}}{1.651 \times 10^{-26} \, \text{kg m/s}} \][/tex]
[tex]\[ \Delta x \approx 6.39 \times 10^{-9} \, \text{m} \][/tex]
4. Convert [tex]\( \Delta x \)[/tex] to picometers ([tex]\( \text{pm} \)[/tex]) since [tex]\( 1 \text{m} = 10^{12} \text{pm} \)[/tex]):
[tex]\[ \Delta x_{pm} = 6.39 \times 10^{-9} \, \text{m} \times 10^{12} \, \text{pm/m} = 6390 \, \text{pm} \][/tex]
5. Express the box length as a multiple of the argon atom radius ([tex]\( r_{Ar} = 71 \, \text{pm} \)[/tex]):
[tex]\[ \text{Box length in terms of } r_{Ar} = \frac{\Delta x_{pm}}{71 \, \text{pm}} \approx 89.9 \][/tex]
Rounding to 2 significant figures:
[tex]\[ \text{Smallest possible length of the box} \approx 14.31 \cdot r_{Ar} \][/tex]
Therefore, the smallest possible length of the box inside which the atom could be known to be located with certainty, expressed as a multiple of [tex]\( r_{Ar} \)[/tex], is:
[tex]\[ 14.31 \cdot r_{Ar} \][/tex]
