Question

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$<h5 > hisenburg uncertanity principle$



An electron has a speed of 4,00,000m/s. If it's velocity is accurate up-to 10% then calculate uncertainty in position of the electron ???????



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Answer 1

Hi there !
Here's the answer :

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Heisenberg Uncertainty Principle
It States that less precisely you measure a particle's motion, the less precisely you can know it's motion. And the more precisely you measure a particle's motion, the less precisely you can know it's position.

According to Heisenberg Uncertainty Principle, the product of Uncertainties in position (∆x) and Velocity (∆v) is always greater than or equal to $\frac{h}{n\pi}$

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¶¶¶POINTS TO REMEMBER:

For an electron,
mass $m = 9.1 × 10^{-31}\: kg$
and n = 4

Planck's constant $h = 6.63×10^{-34}\: Js$

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¶¶¶SOLUTION:

Given,
v = 400000 m/s

and

Velocity is accurate Upto 10%
=> The Uncertainty in velocity is 10% of the actual value, 400000 m/s

¶ Find Uncertainty in velocity (∆v)

Uncertainty in velocity ∆v = $4×10^{5} × \frac{10}{100}$

=> ∆v = $4×10^{4}\: m/s$

¶ Find Uncertainty in position (∆x)

∆x =$\frac{h}{n\pi}$× $\frac{1}{m\: deltav}$

=> ∆x = $\frac{6.63×10^{-34}}{4×3.14×9.1×10^{-31}×4×10^{4}}$

=> ∆x = $\frac{6.63}{4×4×3.14×9.1}×\frac{10^{-34}}{10^{-27}}$

=> ∆x = $1.45×10^{-9}\: m$

=> ∆x = 1.45 nm

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…

Answer 2

∆xm∆v=h/(nπ)

∆x:- uncertainty position
m:- mass of electron
∆v:- velocity×accuracy
h:- Heisenberg constant
n:-4(for electron)
π:-22/7

Now,
$dx \times m \times dv = \frac{h}{4\pi} \\ dx = \frac{6.63 \times {10}^{ - 34} }{4 \times 3.14 \times dv \times m} \\ \\ dv = v \times \frac{10}{100} \\ \\ dv = 4 \times {10}^{4} \\ \\ now \\ dx = \frac{6.63 \times {10}^{ - 34} }{4 \times 3.14 \times 4 \times {10}^{4} \times \times 9.1 \times {10}^{ - 31} } \\ = \frac{6.63 \times {10}^{ - 7} }{16 \times 3.14 \times 9.1} \\ dx = \frac{6.63 \times {10}^{ - 7} }{457.184} \\ \\ dx = 0.01450 \times {10}^{ - 7} m \\ 1.45 \: nm$

HOPE IT WILL HELP YOU‼️