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kompoz [17]
3 years ago
8

An electron has an uncertainty in its position of 587 pm . part a what is the uncertainty in its velocity?

Physics
1 answer:
saul85 [17]3 years ago
6 0
We can solve the problem by using Heisenberg's uncertainty principle, which states that:
\Delta x \Delta p  \geq   \frac{h}{4 \pi}
where
\Delta x is the uncertainty on the position
\Delta p is the uncertainty on the momentum
h is the Planck constant.

Keeping in mind that the momentum is the product between the mass of the electron and its velocity: 
p=mv
we can rewrite the Heisenberg principle as
m \Delta x \Delta v \geq \frac{h}{4 \pi}
where \Delta v is the uncertainty on the velocity.

The uncertainty on the position is \Delta x = 587 pm = 587 \cdot 10^{-12} m, so we can find the uncertainty on the velocity by re-arranging the previous equation:
\Delta v  \geq  \frac{h}{4 \pi m \Delta x}=  \frac{6.6 \cdot 10^{-34} Js}{4 \pi (9.1 \cdot 10^{-31}kg)(587 \cdot 10^{-12} m)} =9.84 \cdot 10^4 m/s

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