Answer:
The uncertainty in its position is 308 pm
Explanation:
Given;
velocity of electron, v = 3.7 x 10⁵ m/s
uncertainty in velocity of the electron, Δv = 1.88 x 10⁵ m/s
Planck's constant, h = 6.626 x 10⁻³⁴ J.s
mass of electron, m = 9.109 x 10⁻³¹ kg
Using Heisenberg's uncertainty principle, we will determine the uncertainty in the position of the electron.
Δx(mΔv) ≥ h/4π
where;
Δx is uncertainty in position
m is mass of the electron
Δv is uncertainty in velocity of the electron
h is Planck's constant
Δx [tex]= \frac{h}{4\pi (mv) } = \frac{6.626*10^{-34}}{4\pi (9.109*1.88*10^5)} = 3.08*10^{-10} \ m[/tex]
Δx = 308 x 10⁻¹² m
Δx = 308 pm
Therefore, the uncertainty in its position is 308 pm
The uncertainty in its position is 308 pm
Since
velocity of electron, v = 3.7 x 10⁵ m/s
uncertainty in velocity of the electron, Δv = 1.88 x 10⁵ m/s
Planck's constant, h = 6.626 x 10⁻³⁴ J.s
mass of electron, m = 9.109 x 10⁻³¹ kg
So,
the uncertainity is
= 6.626 x 10⁻³⁴/ 4π (9.109 * 1.88 *10⁵)
= 3.08 x 10^-12 m
= 308 pm
Therefore, The uncertainty in its position is 308 pm
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