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Leviafan [203]
3 years ago
11

What does the term electron orbital describe? What does the term electron orbital describe? An electron orbital describes a thre

e-dimensional space where an electron can be found 90% of the time. An electron orbital describes the exact distance of an electron from the nucleus. An electron orbital describes the orbit of an electron around the nucleus.
Physics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

An electron orbital describes a three-dimensional space where an electron can be found 90% of the time.

Explanation:

According to Heisenberg's theory we cannot observe the position and velocity of an electron in an orbit, but if they were around the nucleus (in orbit), it would be possible to know its velocity and position, which would be contrary to the principle of Heisenberg So we can say that no electron revolves around a certain orbit around the nucleus, so we can only predict if the electron will be in the right position at the right time.

From there we find two definitions for electron orbital let's see:

  • Orbital is considered the region of space, where each electron spends most of its time.
  • Orbital is considered the region of space that is most likely to find an electron.
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In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed
Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is 87.87 * 10^{-37}.

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= 5.29 * 10^{-11} m\\

velocity=2.9* 10^{6} m/s

Working formula, M=N/A

I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}

T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }

   =15.16 * 10^{-5} s

I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }= 0.10 * 10^{-14}

                     =1 * 10^{-15} C

M=1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}

  =87.87 * 10^{-37}

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

#SPJ4

4 0
1 year ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
3 years ago
a car moves 6 km east, then 4km north. what was the distance covered? What was the total displacement?
Aliun [14]

Answer:

distance= 10 km

displacement= 7.21 km

Explanation:

distance= scalar (only magnitude)

displacement= vector (magnitude & direction)

distance= 6 km+ 4 km

= 10 km

displacement= shortest difference btwn 2 pts=

sqrt( 6^2+4^2)

sqrt(52)

7.21 km

7 0
2 years ago
Which energy conversion takes place when a solar cell is used to light a street lamp?
professor190 [17]

Answer:

SOLAR TO LIGHT ENERGY

Explanation:

4 0
3 years ago
Read 2 more answers
2. What is the accelaration of a<br> 4000 kg airplane that has a force<br> of 24,000 N?
Bond [772]

Answer:

c) 6 m/s²

Explanation:

8 0
3 years ago
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