An electron is located on a pinpoint having a diameter of 2.04 µm. What is the minimum uncertainty in the speed of the electron?
Δx• Δp ≥ ħ / 2
Δx •m •Δv ≥ ħ / 2
Δv ≥ ħ / 2• m •Δx
The minimum uncertainty in speed
Δv = ħ / 2• m •Δx =
=( h / 2 π ) / 2• m •Δx =
= h / 4• π •m •Δx .
Using the diameter of the pinpoint for the uncertainty in position
Δv =6.63•10^-34/4•π•9.1•10^-31•2.04•10 ^-6 =
= 28.37 m/s.
To find the minimum uncertainty in the speed of the electron, we can use the Heisenberg uncertainty principle, which states that the uncertainty in the position (Δx) and the uncertainty in the momentum (Δp) of a particle are related by the equation:
Δx * Δp >= h/(4π)
where h is the Planck's constant (approximately equal to 6.626 x 10^-34 J·s).
Here, we are given the diameter of the pinpoint, which is 2.04 µm. The uncertainty in the position (Δx) can be calculated by dividing the diameter by 2:
Δx = 2.04 µm / 2 = 1.02 µm
To convert the uncertainty in position from micrometers (µm) to meters (m), we need to multiply it by 10^-6:
Δx = 1.02 µm * 10^-6 = 1.02 x 10^-6 m
Now, we can calculate the minimum uncertainty in the speed of the electron using the uncertainty principle:
Δx * Δp >= h/(4π)
Δp >= h/(4π * Δx)
Substituting the values:
Δp >= (6.626 x 10^-34 J·s) / (4π * 1.02 x 10^-6 m)
Δp >= 5.16 x 10^-28 kg·m/s
The minimum uncertainty in the speed of the electron (Δv) can be calculated using the uncertainty in momentum:
Δv = Δp / mass of the electron
The mass of the electron is approximately 9.11 x 10^-31 kg:
Δv = (5.16 x 10^-28 kg·m/s) / (9.11 x 10^-31 kg)
Δv ≈ 5.67 x 10^2 m/s
Therefore, the minimum uncertainty in the speed of the electron is approximately 5.67 x 10^2 m/s.
To determine the minimum uncertainty in the speed of the electron, we can make use of Heisenberg's uncertainty principle, which states that the product of the uncertainties in position and momentum of a particle is always greater than or equal to the reduced Planck's constant (h-bar) divided by 2.
The uncertainty in position (Δx) is given as half of the diameter of the pinpoint since the electron is assumed to be confined within it. Thus, Δx = 2.04 µm / 2 = 1.02 µm = 1.02 × 10^(-6) m.
We can calculate the minimum uncertainty in the speed (Δv) by rearranging the equation of uncertainty principle. The equation is given as:
Δx * Δv >= h-bar / 2
Plugging in the known values:
1.02 × 10^(-6) m * Δv >= h-bar / 2
Now, we need to know the value of h-bar, which is the reduced Planck's constant. The value of h-bar is approximately 1.05 × 10^(-34) J·s.
1.02 × 10^(-6) m * Δv >= (1.05 × 10^(-34) J·s) / 2
Simplifying the equation:
Δv >= (1.05 × 10^(-34) J·s) / (2 * 1.02 × 10^(-6) m)
Calculating the minimum uncertainty in speed:
Δv >= 5.15 × 10^(-29) m/s
Therefore, the minimum uncertainty in the speed of the electron is approximately 5.15 × 10^(-29) m/s.