g You want to make simultaneous measurements of the position and momentum of an electron and a proton that are moving along a straight line. An alternate statement of the uncertainty principle involves relationship between position (Δx) and momentum (Δp) uncertainties in the form . If both of them are located with an uncertainty of 1 × 10-10 m, what is the ratio of uncertainty in the velocity of the electron to that of the proton?

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

with 5 seconds, no, but over time if the police car stays at 65m/s ad the speeding car stays at 60, then obviously yes, but the answer to your question would be, no.

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