Question

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?

Answer 1

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