Exercise 2.61
If the position of the electron is measured within an accuracy of ± nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is × nm, is there any problem in defining this value.
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
Cannot be defined as the actual magnitude is smaller than uncertainity.
Heisenberg's uncertainty principle says you cannot shrink the uncertainty in position and the uncertainty in momentum together below a fixed floor — pinning one down tighter only forces the other one up.The principle is written
\[\Delta x \cdot \Delta p \geq \frac{h}{4\pi}
\]
where \(\displaystyle \Delta x \) is the uncertainty in position, \(\displaystyle \Delta p \) is the uncertainty in momentum, and \(\displaystyle h = 6.626 \times 10^{-34}\ \text{J s} \) is Planck's constant.Convert nm to m before touching the formula — leaving out the factor of \(\displaystyle 10^{-9}\) here throws the whole answer off by a billion.
\[\Delta x = 0.002\ \text{nm} = 0.002 \times 10^{-9}\ \text{m} = 2 \times 10^{-12}\ \text{m}
\]Rearranging for the minimum uncertainty in momentum,
\[\Delta p \geq \frac{h}{4\pi\, \Delta x} = \frac{6.626 \times 10^{-34}\ \text{J s}}{4\pi \times 2 \times 10^{-12}\ \text{m}}
\]Work the denominator first: \(\displaystyle 4\pi \times 2 \times 10^{-12} = 12.566 \times 2 \times 10^{-12} = 2.513 \times 10^{-11}\ \text{m} \). Then\[\Delta p \geq \frac{6.626 \times 10^{-34}}{2.513 \times 10^{-11}}\ \text{kg m s}^{-1} = 2.64 \times 10^{-23}\ \text{kg m s}^{-1}
\](A joule is \(\displaystyle \text{kg m}^2\text{s}^{-2}\), so \(\displaystyle \text{J s}/\text{m} = \text{kg m s}^{-1}\) — a momentum unit, as it should be.)So fixing the electron's position to within $\displaystyle 0.002$ nm forces its momentum to be uncertain by at least \(\displaystyle 2.64 \times 10^{-23}\ \text{kg m s}^{-1} \).Now check whether a momentum built from a position spread of $\displaystyle 0.05$ nm — roughly the size of an atomic orbital — can even be called a definite number. Use the same expression, \(\displaystyle h/(4\pi \Delta x) \), but now with \(\displaystyle \Delta x = 0.05\ \text{nm} = 5 \times 10^{-11}\ \text{m} \):
\[p = \frac{h}{4\pi \times 0.05\ \text{nm}} = \frac{6.626 \times 10^{-34}\ \text{J s}}{4\pi \times 5 \times 10^{-11}\ \text{m}}
\]Denominator: \(\displaystyle 4\pi \times 5 \times 10^{-11} = 12.566 \times 5 \times 10^{-11} = 6.283 \times 10^{-10}\ \text{m} \). Then\[p = \frac{6.626 \times 10^{-34}}{6.283 \times 10^{-10}}\ \text{kg m s}^{-1} = 1.05 \times 10^{-24}\ \text{kg m s}^{-1}
\]The comparison between this \(\displaystyle p\) and the \(\displaystyle \Delta p \) found above is the whole point of the question — not either number on its own.
\[\frac{\Delta p}{p} = \frac{2.64 \times 10^{-23}}{1.05 \times 10^{-24}} \approx 25
\]The uncertainty in momentum, \(\displaystyle 2.64\times10^{-23}\ \text{kg m s}^{-1}\), is about $\displaystyle 25$ times bigger than the momentum value itself, \(\displaystyle 1.05\times10^{-24}\ \text{kg m s}^{-1}\). A quantity whose spread of possible values is $\displaystyle 25$ times its own size has not actually been pinned down to a number — zero, negative values, and values many times larger are all equally consistent with that spread. So yes, there is a problem: once the electron's position is known as precisely as $\displaystyle 0.05$ nm (atomic-orbital scale), its momentum cannot be assigned any definite value at all. This is exactly why an electron in an atom cannot be pictured as moving on a fixed orbit with a definite position and a definite momentum at every instant — the uncertainty principle rules that picture out.Answer: Δp ≥ $\displaystyle 2.64$ × $\displaystyle 10$⁻²³ kg m s⁻¹; the corresponding momentum h/($\displaystyle 4$π × $\displaystyle 0.05$ nm) = $\displaystyle 1.05$ × $\displaystyle 10$⁻²⁴ kg m s⁻¹ is about $\displaystyle 25$ times smaller than this uncertainty, so no definite value can be assigned to it — the uncertainty swamps the quantity itself.