Q.Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity ($\Delta V$) is equal to $\dfrac{1}{4\pi}$ of its velocity ($V$).

Step 1. Heisenberg's uncertainty relation, at its minimum (equality) bound: $\Delta x\cdot\Delta p=\frac{h}{4\pi}$

Step 2. Given: the uncertainty in position equals the particle's de Broglie wavelength, $\Delta x=\lambda$. From $\lambda=h/(mV)$ (using $V$ for the particle's actual velocity, to distinguish it from the uncertainty $\Delta V$), substitute $\Delta x=h/(mV)$.

Step 3. Also substitute $\Delta p=m\,\Delta V$ (momentum uncertainty in terms of velocity uncertainty):

Step 4. The mass $m$ cancels from the left-hand side:

Step 5. Cancel $h$ from both sides and rearrange: …

Substitute both Δx = λ = h/mV and Δp = mΔV into the minimum-uncertainty equality, then simplify a …

• Forgetting that m and h both cancel out completely, which is what makes the final result so clean (ΔV depends only on V). …