) What is the minimum uncertainty in the velocity of an electron that is known to be somewhere between 0.050 nm and 0.10 nm from a proton?
delta x = .05 *10^-9 m
delta x * delta p = (1/2) hbar = .5*10^-34
delta p = (.5*10^-34)/.05*10^-9)
= 10 * 10^-25 kg m/s
mass of electron * delta v = 10*-24
delta v = (10*10^-25)/(9*10^-31)
delta v = 1.1 * 10^6 m/s
To determine the minimum uncertainty in the velocity of the electron, we can use the Heisenberg uncertainty principle, which states that there is an inherent uncertainty in measuring certain pairs of properties of a particle, such as position and velocity. The uncertainty in position multiplied by the uncertainty in velocity is always greater than or equal to a certain value called Planck's constant divided by 4π.
Mathematically, the uncertainty principle is expressed as:
Δx Δv ≥ h / (4π)
Where:
Δx is the uncertainty in position
Δv is the uncertainty in velocity
h is Planck's constant (approximately 6.626 x 10^-34 joule-seconds)
In our case, the electron is known to be located somewhere between 0.050 nm and 0.10 nm from a proton. Since we are given a range of positions, the uncertainty in position (Δx) is half of the difference between the maximum and minimum positions:
Δx = (0.10 nm - 0.050 nm) / 2 = 0.025 nm
Now we can rearrange the uncertainty principle to solve for the uncertainty in velocity (Δv):
Δv ≥ h / (4πΔx)
Plugging in the values:
Δv ≥ (6.626 x 10^-34 J·s) / (4π(0.025 x 10^-9 m))
Calculating this expression will give us the minimum uncertainty in the velocity of the electron.