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andrey2020
2 months ago
14

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

In the classical model, the electron orbits around the nucleus, being held in orbit by the electromagnetic interaction between itself and the protons in the nucleus, much like planets orbit around the sun, being held in orbit by their gravitational interaction. When the electron is in a circular orbit, it must meet the condition for circular motion: The magnitude of the net force toward the center, Fc, is equal to mv 2/r. Given these two pieces of information, deduce the velocity v of the electron as it orbits around the nucleus.
Physics
1 answer:
kicyunya [3.2K]2 months ago
7 0

Result:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

v=\sqrt{k\frac{e^2}{m_e r}}

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

r=5.3\cdot 10^{-11}m

while the mass of the electron is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

By plugging in the values into the formula, we achieve

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diame
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For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

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                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

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. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

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A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of
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The average acceleration is

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Response:

(e) thermal expansion

Clarification:

The density, heat of fusion, and melting temperature of a metal are critical factors to consider when increasing its temperature from room temperature to its melting point. These will dictate the following aspects:

Density: refers to the ratio between a body's mass and the space it occupies in the universe.

Heat of fusion: The enthalpy of fusion or heat of fusion signifies the amount of energy required to cause a mole of an element at its melting point to transition from solid to liquid state, under constant pressure.

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