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3 years ago

How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number mℓ = −1?

Physics

1 answer:

LekaFEV [45]3 years ago

3 0

Answer:

6 electrons

Explanation:

To solve this question lets determine the possible quantum numbers for the principal quantum number n = 4.

For the quantum number l which describes the shape of the orbital, we have the possible values : 0 to n-1.

Thus, for n = 4 the l can assume the values 0, 1, 2, 3 ( 4 possible shapes )

For the angular quantum number, ml, which tell us the orientation in space,we have the values - l to + l.

So lets determine the number of orbials which can have the values -l for n=4

l = 0 ml = 0

l= 1 ml = -1,0,1

l= 2 ml = -2, -1, 0, 1, 2

l= 3 ml = -3,-2,-1,0,1,2,3

So we have three orbital with ml = - 1 and from Pauli´s exclusion principle we can have up two electrons in each orbital. Thus for n= 4 we can have up to 6 electrons with ml = -1

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Murljashka [212]

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Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

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Here $d=0.15, R=15, \rho=1.68(10^{-8})$ .

Substituting numerical values,

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4 years ago

A bike rider was moving to the right at a constant speed. Suddenly the wind then starts blowing against her with 3 N of force to

stiks02 [169]

Answer:

The correct option is D

Explanation:

This question can be better understood when discussed using the Newton's first law of motion which states that an object would continue to move with a uniform speed (in a straight line) unless acted upon by an external force. What happens here (in the question) is that the bike rider would have continued moving at a constant speed (to the right) if not for the opposing force of the wind that acted against her (to the left). <u>This wind/force would cause her speed to reduce or slow down (as posited by the law)</u>.

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3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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VashaNatasha [74]

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Explanation:

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