Question

An electron and a 0.048 O-kg bullet each have a velocity of magnitude 450 m/s, accurate to within 0.010 0%. Within what lower limit could we determine the position of each object along the direction of the velocity? for the electron mm for the bullet m

Answer 1

Solution: Heisenberg uncertainty principle states that it is impossible to measure the momentum and position of a particle simultaneously with arbitrarily high precision." The expression for the uncertainty principle is given below: AxAp 2 .. (1) Here, the uncertainty in position is Ax, uncertainty in momentum is Ap, and the Planck's constant is h. The expression for the momentum is given below: Ap = may Here, the mass is m and the uncertainty in velocity is Av. Substitute mAv for Ap in Equation (1). Ax(mAv) 2 Rearrange the above equation to get Ax as follows: Ax2 Armand Substitute 6.626x10- J-s for h. 9.11x10-31 kg for m, 3.14 for 1, and 0.0100% for Av. Ar> 6.626x10-34.s. TV 4(3.14)(9.11x10*' kg) 0.0100% 4505)) 6.626x10-34 J-s 4(3.14) (9.11x10-kg (0.045 m/s) 21.28x10 m Thus, for bullet it would be. 6.626x10-34 J-S Ar> 4(3.14)(0.0480 kg) 0.0100% +Soms) 100 ) 6.626x10-24 J-5 -4(3.14)(0.0480 kg)(0.045m/s) >2.44x10 m