The Heisenberg uncertainty principle, $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$, where $\Delta x=$ Uncertainty in position, $\Delta p=$ Uncertainty in momentum and $h / 4 \pi=$ constant. As $\Delta x$ is same for electron and helium and $h / 4 \pi$ is a constant, therefore minimum uncertainty in the measurement of the momentum of the helium atom will be same as that of an electron which is $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$.