An electron has a speed of 500m/s with uncertainty of 0.02% what is the uncertainty indicating it's position?
To determine the uncertainty in the electron's position, we can make use of Heisenberg's uncertainty principle, which states that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and momentum, can be known simultaneously.
The uncertainty principle can be mathematically expressed as the product of the uncertainties in position (Δx) and momentum (Δp) being greater than or equal to a constant value, typically denoted as h-bar (ħ):
Δx * Δp ≥ ħ
Given the uncertainty in the electron's speed, we need to relate it to momentum. The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v), according to the equation:
p = m * v
Since we are dealing with an electron, we can use its mass (m) as 9.11 x 10^-31 kg.
Considering the uncertainty in the electron's speed, with a value of 0.02%, we need to calculate the uncertainty in its momentum. To do this, we can multiply the electron's mass (m) by the uncertainty in speed (Δv):
Δp = m * Δv
Plugging in the values:
Δp = (9.11 x 10^-31 kg) * (0.02/100 * 500 m/s)
Now we can use the uncertainty principle to find the uncertainty in position:
Δx * Δp ≥ ħ
Δx * [(9.11 x 10^-31 kg) * (0.02/100 * 500 m/s)] ≥ ħ
Finally, solve for Δx:
Δx ≥ ħ / [(9.11 x 10^-31 kg) * (0.02/100 * 500 m/s)]
Now, ħ is a constant equal to approximately 1.05457182 x 10^-34 Js, and by plugging it into the equation along with the given values, you can calculate the uncertainty in the electron's position.