Question

An electron and a 0.0280-kg bullet each have a velocity of magnitude 450 m/s, accurate to within 0.0100%. Within what lower limit could we determine the position of each object along the direction of the velocity?

Answer 1

`8.1466mm\ and\ 2.646* 10^(-31)m`

Step-by-step explanation:

Mass of bullet = 0.028 kg

Velocity v = 450 m/sec

Accuracy in velocity = 0.01 %

So
`\Delta v_x=0.01* (450)/(100)=0.045m/sec`

From the uncertainty principle
`\Delta x\geq (h)/(2\Delta p_x)`

`\Delta x\geq (h)/(2m\Delta v_x )`

For electron
`m=9.1* 10^(-31)kg`

`\Delta x=(6.67* 10^(-34))/(2* 9.1* 10^(-31)* 0.045)=8.144* 10^(-3)m=8.144mm`

For bullet mass m =0.028 kg

`\Delta x=(6.67* 10^(-34))/(2* 0.028* 0.045)=2.646* 10^(-31)m`