HW Question 1.B.25

[605379296]

For this problem, I got delta p as greater than or equal to 1.51*10^-25. However, when I try to find delta v, the solution manual states to multiply by 1/2, so it looks like this: .5((1.5*10^-34)/(9.1*10^-31)(350*10^-12))= 1.65*10^5 m/s. Again, why would you multiply that by .5? Isn't the original formula just [delta v= delta p/m]?

[William Chan 1D]

The equation is ΔpΔx ≥ 1/2 ℏ.

Not that ℏ is equal to h/2π, so in essence, ΔpΔx is equal to h /π, or 1/2 ( h/2π ). I think this is where the other 1/2 came from.

[205150314]

For this problem, I got delta p as greater than or equal to 1.51*10^-25. However, when I try to find delta v, the solution manual states to multiply by 1/2, so it looks like this: .5((1.5*10^-34)/(9.1*10^-31)(350*10^-12))= 1.65*10^5 m/s. Again, why would you multiply that by .5? Isn't the original formula just [delta v= delta p/m]?

Where did the 350*10^m/s come from?

[Paul Hage 2G]

The 350 x 10^-12m/s came from the conversion from picometers to meters. 1pm=10^-12m.