Question 12 Marks
The uncertainty in the position of a moving bullet of mass 10 g is $\Delta x m \Delta v =\frac{ h }{4 \pi}$. Calculate the uncertainty in its velocity?
Answer
View full question & answer→According to the uncertainty principle,
$\Delta x . m \Delta v =\frac{ h }{4 \pi}$ or $\Delta v =\frac{h}{4 \pi m \Delta x} h=6.625 \times 10^{-34} kg m ^2 s^{-1} m=10 g=10^{-2} K$
$
\Delta x=10^{-5} m ; \Delta v=\frac{\left(6.626 \times 10^{-34} kgm^2 s^{-1}\right)}{4 \times 3.143 \times\left(10^{-2} kg\right) \times\left(10^{-5} m\right)}=5.27 \times 10^{-28} mv
$
$\Delta x . m \Delta v =\frac{ h }{4 \pi}$ or $\Delta v =\frac{h}{4 \pi m \Delta x} h=6.625 \times 10^{-34} kg m ^2 s^{-1} m=10 g=10^{-2} K$
$
\Delta x=10^{-5} m ; \Delta v=\frac{\left(6.626 \times 10^{-34} kgm^2 s^{-1}\right)}{4 \times 3.143 \times\left(10^{-2} kg\right) \times\left(10^{-5} m\right)}=5.27 \times 10^{-28} mv
$
