Unformatted text preview: ive direction.
The conservation of linear momentum gives rise to Equation 29.7, which relates the
difference λ ′ − λ between the scattered and incident Xray photon wavelengths to the
scattering angle θ of the electron as λ′ − λ = b h
1 − cos θ
mc h
1
g= mc b− cos 180.0° g= 2h
mc (1) The conservation of total energy is written as
hc λ
4 Energy
of incident
photon + 0
{
Initial
kinetic
energy
of electron = hc
λ′ 4 Energy
of scattered
photon + mv 2
123
1
2 Final
kinetic
energy
of electron (2) Chapter 29 Problems 1543 Equations (1) and (2) will permit us to find the wavelength λ of the incident Xray photon.
SOLUTION Solving Equation (1) for λ ′ and substituting the result into Equation (2) gives
hc λ = hc
+
2h
+λ
mc 1
2 mv 2 Algebraically rearranging this result, we obtain a quadratic equation for λ: λ2 + 2
Fh I
GJ
H3
mc K
1
2
4 .85 × 10 −12 m λ− 2h2 c h=0 1
m 2 mv 2
14243 9 .70 × 10 −20 m 2 where we have used h = 6.63 × 10–34 J⋅s, m = 9.11 × 10–31 kg, c = 3.00 × 108 m/s, and
v = 4.67 × 106 m/s. Solving this quadratic equation for λ, we obtain λ = 3.09 × 10 −10 m CHAPTER 30 THE NATURE OF THE ATOM
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) The Bohr model deals with a single negatively charged electron in orbit about a positively
charged nucleus. Since only one electron is assumed to be present, the model does not take
into account the electrostatic repulsion between electrons that would exist in a multipleelectron atom, such as helium (two electrons).
2. (c) According to Equation 30.10, the radius rn of an orbit is proportional to the square of the
quantum number n. An increase in the radius by a factor of four, therefore, means that n has
doubled. According to Equation 30.8, the magnitude Ln of the orbital angular momentum is
proportional to n. This means that Ln doubles when rn doubles.
3. (b) The shortest wavelength λshortest in a given series can be calculated from Equation 30.14 1
1
RZ 2
= 0 ) as
= 2 , where R is the Rydberg constant and Z is the number of
λshortest
ni2
nf
protons in the nucleus. (with 4. 1282 nm
5. (e) To remove an electron completely from an atom and place it at rest infinitely far from the
nucleus of the atom, energy must be supplied in the amount of the ionization energy.
6. (e) In the quantum mechanical picture, the principal quantum number n can have the values
1, 2, 3, …. The orbital quantum number l can have the values l = 0, 1, 2, ..., ( n − 1) . Thus,
for n = 3, the possible values for l are 0, 1, and 2.
7. (a) According to Equation 30.15, the magnitude L of the orbital angular momentum is
h
L = l ( l + 1)
. This expression can be solved for l, the orbital angular momentum
2π
quantum number, with the result that l = 4. Since l can be at most n − 1, the principal
quantum number n must be 5 or larger. Of the options given in this case, 5 is the only
feasible answer.
8. (c) The spin angular momentum of the electron plays no role in the Bohr model, but it does
play a role in the quantum mechanical picture.
9. 6.33 × 10−34 J·s
10. (e) The Pauli exclusion principle states that no two electrons in an atom can have the same
set of the four quantum numbers n, l, ml, and ms. Chapter 30 Answers to Focus on Concepts Questions 1545 11. (a) A subshell can contain up to a maximum of 2(2l + 1) electrons, where l is the orbital
quantum number. For the 5f subshell, l = 3, so the maximum number of electrons is
2(2l + 1) = 14. For the 6h subshell, l = 5, so the maximum number of electrons is
2(2l + 1) = 22. Therefore, 19 electrons can fit into the 6h subshell but not the 5f subshell.
12. (c) The maximum number is obtained by adding the maximum number for each subshell
within the n = 5 shell. The value of the orbital quantum number l defines a subshell, and for
n = 5, the possible values for l are 0, 1, 2, 3, and 4. Each subshell can contain up to a
maximum of 2(2l + 1) electrons. Thus, the l = 0 subshell can hold 2 electrons, the l = 1
subshell can hold 6 electrons, the l = 2 subshell can hold 10 electrons, the l = 3 subshell can
hold 14 electrons, and the l = 4 subshell can hold 18 electrons, for a total of 50.
13. (d) For the s, p, and d subshells the values for l are 0, 1, and 2, respectively. Each subshell
can contain up to a maximum of 2(2l + 1) electrons. Thus, the s subshell can hold 2
electrons, the p subshell can hold 6 electrons, and the d subshell can hold 10 electrons. In
addition, the order in which the subshells fill is important. The subshells fill in order of
increasing principal quantum number n and, for a given n, in order of increasing l. There are
exceptions, however. For example, the 4s subshell (l = 0) fills before the 3d subshell (l = 2).
This answer shows the reverse, namely, that the 3d subshell fills before the 4s subshell.
14. (c) More energy is required to knock a Kshell electron completely out of an atom, when the
number Z of protons in the nucleus is greater. This is because the electrostatic force of
attraction that the nucleus exerts on an electro...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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