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Tamiku [17]
3 years ago
15

The quantum state of a particle can be specified by giving a complete set of quantum numbers (n,l, ml,ms). how many different qu

antum states are possible if the principal quantum number is n = 2?
Mathematics
1 answer:
ipn [44]3 years ago
7 0

To find the total number of allowed states, first, write down the allowed orbital quantum numbers l<span>, and then write down the number of allowed values of </span>m<span>l for each orbital quantum number. Sum these quantities, and then multiply by 2 to account for the two possible orientations of spin.</span>

Quantum states for n=5
l......ml
0 , 0................................1
1 , -1, 0, +1................................ 3
2 , -2, -1, 0, +1, +2................................5
3 , -3, -2, -1, 0, +1, +2, +3................................ 7
4 , -4,-3, -2, -1, 0, +1, +2, +3,+4 ................................9

2*(1+3+5+7+9)=
2*5^2 = 50 allowed states

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Which function has zeros at x = -2 and x = 5?
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Answer:

<em>f(x)=x²-3x-10</em>

Step-by-step explanation:

\begin{gathered}f(x) = x {}^{2} - 3x - 10 \\ to \: find \: x \: intercept \:o r \: zero \: substitute \: f(x) = 0\: \\ 0 = x {}^{2} - 3x - 10 \\ x {}^{2} - 3x - 10 = 0 \\ x {}^{2} + 2x - 5x - 10 = 0 \\ x(x + 2) - 5x - 10 = 0 \\ x(x + 2) - 5(x + 2) = 0 \\ (x + 2).(x - 5) = 0 \\ x + 2 = 0 \\ x - 5 = 0 \\ x = - 2 \\ x = 5\end{gathered}

f(x)=x

2

−3x−10

tofindxinterceptorzerosubstitutef(x)=0

0=x

2

−3x−10

x

2

−3x−10=0

x

2

+2x−5x−10=0

x(x+2)−5x−10=0

x(x+2)−5(x+2)=0

(x+2).(x−5)=0

x+2=0

x−5=0

x=−2

x=5

therefore the zeros of the equation are x₁=-2,x₂=5

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Write an equation that describes the function.
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4x is the proportionality
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Rich is building a travel crate for his dog, Thomas, a beagle-mix who is about 28 inches long, 11 inches wide, and 24 inches tal
Dmitry [639]

Answer:

27600

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re
fiasKO [112]

Answer: R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts 38^{\circ} angle to each other

The resultant of the two forces is given by

\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}

Insert the values

\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb

Resultant makes an angle of

\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}

So, the resultant makes an angle of 22.57^{\circ} with 9 lb force

Angle made with 13 lb force is 38^{\circ}-22.57^{\circ}=15.43^{\circ}

7 0
3 years ago
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