Question

During the period of time required for light to pass through a hydrogen atom \(\left(r=0.53 \cdot 10^{-10} \mathrm{~m}\right),\) what is the least uncertainty in the atom's energy? Express your answer in electron-volts.

Short answer

Answer: The least uncertainty in energy for the hydrogen atom during the time it takes for light to pass through it is approximately 937.5 electron volts.

Step-by-step solution

Understand the Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle states that there is a fundamental limit to the precision with which certain pairs of physical variables can be known simultaneously. One of the key expressions of this principle is given by the following formula: \(\Delta E \cdot \Delta t \geq \frac{h}{4 \pi}\) Here, \(\Delta E\) is the uncertainty in energy, \(\Delta t\) is the uncertainty in time, and \(h\) is the Planck's constant (\(h = 6.63 \times 10^{-34} \mathrm{Js}\)). In our problem, we are asked to find the least uncertainty in energy, which means finding \(\Delta E\) when the inequality becomes an equality (\(\Delta E \cdot \Delta t = \frac{h}{4 \pi}\)).

Calculate the time it takes for light to pass through the hydrogen atom

We are given the radius of the hydrogen atom (\(r = 0.53 \times 10^{-10} \mathrm{m}\)), and we want to find the time it takes for light to pass through. We can use the speed of light (\(c = 3.0 \times 10^8 \mathrm{\frac{m}{s}}\)) and the formula for time taken: \(\Delta t = \frac{2r}{c}\) (we use 2r instead of r since we are considering the time taken for light to pass through the entire diameter of the atom) \(\Delta t = \frac{2 (0.53 \times 10^{-10} \mathrm{m})}{3.0 \times 10^8 \mathrm{\frac{m}{s}}}\) \(\Delta t \approx 3.53 \times 10^{-18}\) s

Calculate the least uncertainty in energy

Now that we have the uncertainty in time, we can use the Heisenberg Uncertainty Principle to find the least uncertainty in energy: \(\Delta E \cdot \Delta t = \frac{h}{4 \pi}\) \(\Delta E = \frac{h}{4 \pi \cdot \Delta t} = \frac{6.63 \times 10^{-34} \mathrm{Js}}{4 \pi \cdot 3.53 \times 10^{-18} \mathrm{s}}\) \(\Delta E \approx 1.5 \times 10^{-16} \mathrm{J}\)

Convert the uncertainty in energy to electron volts

Finally, we need to convert the uncertainty in energy from joules to electron volts. We know that 1 electron volt (eV) is equal to \(1.6 \times 10^{-19} \mathrm{J}\): \(\Delta E (\mathrm{eV}) = \frac{\Delta E (\mathrm{J})}{1.6 \times 10^{-19} \mathrm{J/eV}}\) \(\Delta E (\mathrm{eV}) = \frac{1.5 \times 10^{-16} \mathrm{J}}{1.6 \times 10^{-19} \mathrm{J/eV}}\) \(\Delta E (\mathrm{eV}) \approx 937.5\) eV The least uncertainty in energy for the hydrogen atom during the time it takes for light to pass through it is approximately 937.5 electron volts.