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SpyIntel [72]
2 years ago
12

Suppose an electron is trapped within a small region and the uncertainty in its position is 24. 0 x 10-15 m. what is the minimum

uncertainty in the electron's momentum?
Physics
1 answer:
sergeinik [125]2 years ago
3 0

The minimum uncertainty in the electron's momentum is

Δp = 2.2822 \times 10 {}^{ - 20} kg/ms

Given:

Uncertainty in position (ΔX)

= 24 \times 10 {}^{ - 15}m

planck's constant (h)

= 6.26 \times 10 {}^{ - 34} js

To find:

uncertainty in momentum (Δp)

Δx.Δph/4π

24 \times 10 {}^{ - 15} .Δp =  \frac{6.26 \times 10 {}^{ - 34} }{4 \times  \frac{22}{7} }

24 \times 10 {}^{ - 15}. Δp =  \frac{6.26 \times 10 {}^{ - 34} \times 7 }{8}

Δp =  \frac{43.82 \times 10 {}^{ - 34} }{8 \times 24 \times 10 {}^{ - 15} }

Δp =  \frac{43.82 \times 10 {}^{ - 34}  \times 10 {}^{ 15} }{192}

Δp =  \frac{4382 \times 10 {}^{ - 21} }{192}

Δp =22.822 \times 10 {}^{ - 21}

Δp = 2.2822  \times 10 {}^{ - 20} kg/ms

learn more about electron's momentum from here: brainly.com/question/28203580

#SPJ4

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