# What unit does $\Delta x$ have in the uncertainty principle?

Can somebody tell me how the units work out in Heisenberg's principle equation? Mass in $kg$ and velocity in $m/s$ cancel partially with Planck's constant, so what kind of unit is given to $Δx$ to balance the units?

$(m) \cdot (kg \cdot \frac{m}{s}) \ge J \cdot s = (kg \cdot \frac{m^2}{s^2}) \cdot s$
So, the unit is $kg \cdot \frac{m^2}{s}$ on both sides.
The position-momentum uncertainty relation is: $\Delta x\Delta p\geq \frac{\hbar}{2}$. Here $\Delta x$ is the 'uncertainty' aka standard deviation of observing a quantum particle at a given point. The standard deviation is defined as $\Delta x = \sqrt{\left\langle x^2\right\rangle-\left\langle x\right\rangle^2}$. Here x has the dimensions of length and we can then see that $\Delta x$ also has the dimensions of a length. As the uncertainty in momentum is defined in much the same way we see that $\Delta p$ has the dimensions of momentum: $\frac{\text{mass}\cdot\text{length}}{\text{time}}=$. Now $\hbar$ has to have the dimension $\frac{\text{mass}\cdot\text{length}^2}{\text{time}}=\text{energy}\cdot\text{time}$ or else we wouldn't have a physically or mathematically sensible inequality.