An electron in a Carbon atom (number of protons = Z) has the following wave function Zr 1 Zr 1 3 (r,0,0) = (22₂) ·(2) ³. # ( ½ 2² (¹-2) ¹ + 13 (7) e 200 :2 VATT Cos 0 √2√√3 4π What is the probability that this electron is found in the state a) n = 1,1 = 0, m = 0 b) n = 2,1 = 0, m = 0 c) n = 2,1=1,m=1 d) n = 2,1 = 1, m = 0 e) n = 2,1 = 1, m = -1

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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An electron in a Carbon atom (number of
protons = Z) has the following wave function
Zr 1
Zr 1
3
4(r.0, 4) = (22)
·(2) ³. # ( ½ 2² (¹-2) ¹ + 1 § (7)
e 200 :2
Cos 0
VAT
4π
What is the probability that this electron is
found in the state
a) n = 1,1 = 0, m = 0
b) n =
2,1 = 0, m = 0
c) n =
2,1 =
1, m=1
d) n =
2,1 =
1, m = 0
e) n = 2,1 =
1, m = -1
Transcribed Image Text:An electron in a Carbon atom (number of protons = Z) has the following wave function Zr 1 Zr 1 3 4(r.0, 4) = (22) ·(2) ³. # ( ½ 2² (¹-2) ¹ + 1 § (7) e 200 :2 Cos 0 VAT 4π What is the probability that this electron is found in the state a) n = 1,1 = 0, m = 0 b) n = 2,1 = 0, m = 0 c) n = 2,1 = 1, m=1 d) n = 2,1 = 1, m = 0 e) n = 2,1 = 1, m = -1
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