According to Heisenberg's uncertainty principle, $\Delta x \cdot \Delta p \geq$:
A$\dfrac{h}{2\pi}$
B$h$
C$\dfrac{h}{4\pi}$
D$\dfrac{h}{4}$
Answer & Solution
Correct answer: C. $\dfrac{h}{4\pi}$
$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} = \dfrac{\hbar}{2}$, where $\hbar = h/2\pi$ is the reduced Planck constant.
**Meaning:** you cannot simultaneously know an electron's position and momentum with arbitrary precision.
Trying to pin one down makes the other less knowable.
This forces us to talk about probability clouds (orbitals) rather than precise electron trajectories.
Option B is often a near miss in memory because $\hbar = h/2\pi$, but the uncertainty bound has the factor of 2 baked into the denominator twice over.
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