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Makovka662 [10]
3 years ago
9

What is the only possible value of ml for an electron in an s orbital?

Physics
1 answer:
Archy [21]3 years ago
4 0

Answer:

  • zero

Explanation:

m_l     is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

  • n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

  • l : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

       This quantum number is related to the type (or shape) of the orbital:

        For s orbitals l=0

        For p orbitals l=1

        For d orbitals l=2

         For f orbitals l=3

In this case, it is an s orbital, so we have l=0.

  • m_l , the third quantum number can have integer values  {from-l}   to    {+l}

       Since, for the s orbitals  l=0 , the only possible value for {m_l} is zero.

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X axis - time
Y - distance
4 0
2 years ago
A jet makes a landing traveling due east with a speed of 120 m/s .
vesna_86 [32]

Average acceleration over a time interval lasting \Delta t is

a_{\rm ave}=\dfrac{\Delta v}{\Delta t}

where \Delta v is the difference in the jet's final and initial velocities. It's coming to a rest, so

a_{\rm ave}=\dfrac{0-120\frac{\rm m}{\rm s}}{13.5\,\rm s}=-8.9\dfrac{\rm m}{\mathrm s^2}

so the average acceleration has magnitude 8.9 m/s^2 and is pointing West (the direction opposite the jet's movement, which should make sense because the jet is slowing down).

7 0
3 years ago
A man stands on a platform that is rotating at 3.8 rpm; his arms are outstretched and he holds a brick in each hand. The rotatio
Ierofanga [76]

Answer:

a) 5.197rev/s

b) Kf/Ki =2.28

Explanation:

a) Angular momentum of the system L = Iw

ButLi=Lf

Kiwi =Ifwf

wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s

b)Kinetic energy KE= 0.5Iw^2

Ki = 0.5Iiwi^2

Kf=0.5Ifwf^2

Kf/Ki = Ifwf/Iiwi

Kf/Ki = (4.65/3.4))(5.197/3.8)

Kf/Ki = 1.22(1.368)^2

Kf/Ki = 2.28

8 0
3 years ago
Read 2 more answers
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
In all atoms of bismuth the number of electrons must equal to
Viefleur [7K]

Answer:

number of protons

Explanation:

3 0
2 years ago
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