for an atom of mercury, an electron in the 1s shell has a velocity of about 58% of the speed of light. If the mass of the electron at such speed is 1.23 Me and the uncertainty in velocity is 1000 m/s, what is the uncertainty in position of this electron
To determine the uncertainty in the position of the electron, we will use Heisenberg's Uncertainty Principle, which states that the product of the uncertainties in position and momentum of a particle is greater than or equal to Planck's constant divided by 4π.
The uncertainty in momentum (Δp) can be calculated using the mass and uncertainty in velocity provided:
Δp = m * Δv
Here, the mass (m) of the electron is given as 1.23 Me (where Me is the mass of an electron), and the uncertainty in velocity (Δv) is 1000 m/s.
To find the uncertainty in position (Δx), we rearrange the uncertainty principle equation as follows:
Δx * Δp >= h / (4π)
Now we can substitute the values into the equation:
Δx * (m * Δv) >= h / (4π)
Plugging in the given values:
Δx * (1.23 Me * 1000 m/s) >= h / (4π)
To proceed further, we need to convert the mass of the electron from units of Me to kg. Since 1 Me = 9.10938356 × 10^(-31) kg, the mass of the electron can be written as follows:
m = 1.23 Me = 1.23 * 9.10938356 × 10^(-31) kg
Substituting this value, along with Planck's constant (h = 6.62607015 × 10^(-34) J·s) and the value of π, we get:
Δx * (1.23 * 9.10938356 × 10^(-31) kg * 1000 m/s) >= (6.62607015 × 10^(-34) J·s) / (4 * 3.14)
Now, solving for Δx:
Δx >= (6.62607015 × 10^(-34) J·s / (4 * 3.14)) / (1.23 * 9.10938356 × 10^(-31) kg * 1000 m/s)
Evaluating the expression:
Δx >= 5.383 × 10^(-11) meters
Therefore, the uncertainty in the position of the electron is approximately 5.383 × 10^(-11) meters.