Respuesta :
Uncertainty in position:
If the electron has an uncertainty in its velocity of 1.40 m/s then the uncertainty in its position is [tex]4.14\times10^{-4} \text{ m}[/tex].
Heisenberg's uncertainty principle to calculate the required:
Step-1:
We have to apply Heisenberg's uncertainty principle to calculate the uncertainty in the position of an electron. According to the principle:
[tex]$\Delta x \Delta p \geq \frac{h}{4 \pi}$[/tex]
Here,
[tex]$\Delta x$[/tex] is the uncertainty on the position measurement
[tex]$\Delta p$[/tex] is the uncertainty on the momentum measurement
h is the Planck constant.
It is known that the momentum of a particle is calculated as, the product of the mass of the particle and its velocity.
Therefore,
p=mv
Thus the Heisenberg principle becomes:
[tex]$m \Delta x \Delta v \geq \frac{h}{4 \pi}$[/tex]
Here [tex]$\Delta v$[/tex] is the uncertainty in the velocity measurement.
It is given that, [tex]$\Delta v$[/tex]=1.40 m/s. The particle is electron here, thus the mass of the particle m=[tex]9.1 \cdot 10^{-31} \mathrm{~kg}[/tex] and the value of the Plank's constant is, h=[tex]6.62 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} / \mathrm{s}[/tex]
Step-2:
Substituting the values into the equation to get the value of the position uncertainty.
[tex]\Delta x \geq \frac{h}{4 \pi\times m\times \Delta v}\\\geq\frac{6.62\times10^{-34}}{4\pi \times 9.1\times10^{-31}\times1.40} \text{ m}\\\geq 4.14\times10^{-4} \text{ m}[/tex]
To know more about the uncertainty in position, refer to:
https://brainly.com/question/9574825
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