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Find the uncertainty in the position of an electron moving with a velocity of 3 x 10^4 cm/sec accurate up to 0.011%?
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Find the uncertainty in the position of an electron moving with a velo...
Calculating Uncertainty in Electron Position:

To calculate the uncertainty in the position of an electron, we need to use the Heisenberg uncertainty principle, which states that the product of the uncertainty in position and the uncertainty in momentum must be greater than or equal to Planck's constant divided by 4π.

Heisenberg Uncertainty Principle:

ΔxΔp ≥ h/4π

Where, Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant (6.626 x 10^-34 J·s).

Calculating Uncertainty in Momentum:

We can calculate the uncertainty in momentum using the formula:

Δp/p = accuracy/100

Where, p is the momentum of the electron, and accuracy is the given accuracy of the velocity.

Δp/p = 0.011/100 = 0.00011

So, the uncertainty in momentum is 0.00011 times the momentum of the electron.

Calculating Uncertainty in Position:

Now, we can use the Heisenberg uncertainty principle to find the uncertainty in position:

ΔxΔp ≥ h/4π

Δx * 0.00011p ≥ 6.626 x 10^-34 J·s/4π

Δx * 0.00011 * m * v ≥ 6.626 x 10^-34 J·s/4π

Where, m is the mass of the electron, and v is the velocity of the electron.

Substituting the given values, we get:

Δx * 0.00011 * 9.11 x 10^-31 kg * 3 x 10^4 cm/s ≥ 6.626 x 10^-34 J·s/4π

Simplifying the equation, we get:

Δx ≥ 3.08 x 10^-10 cm

Therefore, the uncertainty in the position of the electron is 3.08 x 10^-10 cm.

Conclusion:

The uncertainty in the position of an electron moving with a velocity of 3 x 10^4 cm/sec accurate up to 0.011% is 3.08 x 10^-10 cm. This uncertainty arises due to the Heisenberg uncertainty principle, which states that the product of the uncertainty in position and the uncertainty in momentum must be greater than or equal to Planck's constant divided by 4π.
Community Answer
Find the uncertainty in the position of an electron moving with a velo...
∆x•∆p≥h/4π=5.27×10^-35 js,
∆x•m∆v≥h/4π
therefore ∆x=1.9×10^-7m
& accurate up to.011%~2.09×10^-7m
Note:∆v is taken in m/s . I hope you can understand from the above hints.
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