Answer to Question #137259 in Quantum Mechanics for maria

Question

In a measurement of position and momentum that involved an uncertainty of 0.003 %, the speed of an electron was found to be 800 m/s. Calculate the corresponding uncertainty that arises in determining its position. *

Answer 1

solution

uncertainty in measurement=0.003 %

speed of electron(v)=800 m/s

uncertainty in speed($\Delta$ v) =$\frac{0.003}{100}\times800\\$

=0.024 m/s

according to uncertainty principle

$\Delta x.\Delta p \ge \frac{h}{4\pi}$

$\Delta x.m\Delta\text{ v} \ge \frac{h}{4\pi}$

and

$m=9.1\times10^{-31}kg\\h=6.6\times10^{-34}J-s$

$\Delta x=\frac{h}{4\pi m\Delta \text{v}}$

$\Delta x=\frac{6.6\times10^{-34}}{4\pi \times9.1\times 10^{-31}\times 0.024}$

$\Delta x=0.0024 m$

so uncertainty in position is 0.0024 m .