An electron traveling at 3.1 x 10^5 m/s has an uncertainty in it's velocity of 1.07 x 10^5 m/s. What is the uncertainty in it's position?
delta (mv) delta (position) = hbar/2pi = h/4pi
delta (mv) = m electron (1.07*10^5)
To find the uncertainty in the position of the electron, we can use the Heisenberg Uncertainty Principle, which states that there is a fundamental limit to the precision with which certain pairs of physical properties, such as position and velocity, can be known simultaneously.
The Heisenberg Uncertainty Principle is expressed as:
Δx * Δp ≥ h/4π
where Δx is the uncertainty in position, Δp is the uncertainty in momentum (which is equal to the mass of the electron times the uncertainty in velocity), h is Planck's constant (6.626 x 10^-34 J*s), and π is a mathematical constant approximately equal to 3.14159.
Given that the electron's velocity has an uncertainty of 1.07 x 10^5 m/s, we can calculate the uncertainty in momentum:
Δp = m * Δv
where m is the mass of the electron (approximately 9.11 x 10^-31 kg) and Δv is the uncertainty in velocity.
Putting all the values into the equation:
Δx * Δp ≥ h/4π
Δx * (m * Δv) ≥ h/4π
Δx ≥ h/(4π * m * Δv)
Plugging in the values:
Δx ≥ (6.626 x 10^-34 J*s) / (4π * (9.11 x 10^-31 kg) * (1.07 x 10^5 m/s))
Evaluating this expression will give us the uncertainty in position (Δx).