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ololo11 [35]
3 years ago
8

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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Answer:

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c) H=294300\ m

d) t_T=544.95\ s

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

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<u>For the course of upward acceleration:</u>

using eq. of motion,

s_a=ut+\frac{1}{2}at_a^2

where:

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s_a= height the rocket travels just before its fuel finishes off

so,

s_a=0+\frac{1}{2}\times 19.62\times 100^2

a) s_a=98100\ m is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

v_a=u+at_a

v_a=0+19.62\times 100

b) v_a=1962\ m.s^{-1} is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

v^2=v_a^2-2gh

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g= acceleration due to gravity

v= final velocity of the rocket at the top height

0^2=1962^2-2\times 9.81\times h

h=196200\ m

c) So the total height at which the rocket gets:

H=h+s

H=196200+98100

H=294300\ m

d)

Time taken by the rocket to reach the top height after the fuel is over:

v=v_a+g.t

0=1962-9.81t

t=200\ s

Now the time taken to fall from the total height:

H=v.t'+\frac{1}{2}\times gt'^2

294300=0+0.5\times 9.81\times t'^2

t'=244.95\ s

Hence the total time taken by the rocket to strike back on the earth:

t_T=t_a+t+t'

t_T=100+200+244.95

t_T=544.95\ s

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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Answer:

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