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Unformatted text preview: 44 CHAPTER 4. SINGLEPARTICLE SYSTEMS where a0 = 4πǫ0 h2 /me e2 , to ﬁnd the ground state energy. Express in eV, where 1 eV equals
¯
1.602,2 10−19 J. Values for the physical constants can be found at the start of this section and
in the notations section.
Answer: First verify the Bohr radius
a0 = 4π 8.854 10−12 C2 /J m (1.054,6 10−34 J s)2
= 0.529,2 10−10 m
9.109 10−31 kg (1.602,2 10−19 C)2 Next, taking n = 1 for the ground state,
E1 = − 1
1 eV
(1.054,6 10−34 J s)2
= −2.179,9 10−18 J
= −13.61eV
−31 kg(0.529,2 10−10 m)2 12
2 9.109 10
1.602,2 10−19 J 4.2.3 Discussion of the eigenvalues 4.2.3.1 Solution hydca Question: If there are inﬁnitely many energy levels E1 , E2 , E3 , E4 , E5 , E6 , . . ., where did they
all go in the energy spectrum?
Answer: The En for large values of n are all graphically indistinguishable from zero energy.
The electron is almost free from the nucleus in those energy states. 4.2.3.2 Solution hydcb Question: What is the value of energy level E2 ? And E3 ?
Answer: No need to put all the numbers into
h2 1
¯
En = −
2me a2 n2
0
1
because the only diﬀerence between E2 and E1 is just a ﬁnal factor 4 . So E2 = 1 E1 = −3.4
4
eV. Similarly, E3 = 1 E1 = −1.51 eV.
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This note was uploaded on 01/06/2012 for the course PHY 3604 taught by Professor Dr.danielarenas during the Fall '11 term at UNF.
 Fall '11
 Dr.DanielArenas
 mechanics, Energy

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