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In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.10 nm. If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving

Respuesta :

Answer:

[tex]\Delta v = 5.8 \times 10^5 m/s[/tex]

Explanation:

given

uncertainly in the position of the electron  = 0.10 n m  ([tex]\Delta x)[/tex]

using uncertainly principle

[tex]\Delta x .\Delta p = \frac{h}{4 \pi }[/tex]

[tex]m\Delta v.\Delta x = \frac{h}{4 \pi }\\[/tex]

after solving these equation we get

[tex]\Delta v = \frac{h}{4 \pi m\Delta x}[/tex]

[tex]\Delta v = \frac{6.6\times 10^{34}} { 4 \times 3.14 \times 9.1\times 10^{-31} \times 0.10 \times 10^{-9}}[/tex]

after solving we get

[tex]\Delta v = 5.8 \times 10^5 m/s[/tex]

hamzaahmeds

This question involves the concepts of uncertainty principle and momentum.

The speed of electron is "5.8 x 10⁵ m/s".

What is the Uncertainty Principle?

The uncertainty principle states that:

[tex]\Delta P \Delta x=\frac{h}{4\pi}\\\\m\Delta v\Delta x =\frac{h}{4\pi} \\\\\Delta v = \frac{h}{4\pi m\Delta x}[/tex]

where,

  • Δv = uncertainty in speed = speed = ?
  • m = mass of electron = 9.1 x 10⁻³¹ kg
  • Δx = uncertainty in position = 0.1 nm = 1 x 10⁻¹⁰ m
  • h = plank's constant = 6.63 x 10⁻³⁴ J.s

Therefore,

[tex]\Delta v = \frac{6.63 x 10^{-34}\ J.s}{4\pi (9.1\ x\ 10^{-31}\ kg)(1\ x\ 10^{-10}\ m)}[/tex]

Δv = 5.8 x 10⁵ m/s

Learn more about uncertainty principle here:

https://brainly.com/question/13050159

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