Unformatted text preview: y 4. Because the
atomic mass number of the daughter nucleus, lead 207 Pb , is 207, we have that
82 A − 4 = 207 or A = 207 + 4 = 211 Similarly, given that the atomic number of the daughter nucleus is 82, the atomic number Z
of the parent nucleus must be Z = 82 + 2 = 84. This atomic number identifies the parent
nucleus as polonium (Po). Therefore, the decay process is 211
207
4
84 Po → 82 Pb + 2 He . b. In a β − decay, the atomic mass number A is unaffected, but the atomic number Z
increases by 1. Therefore, the parent nucleus in this decay process has an atomic number
that is 1 less than that of the daughter nucleus: Z = 82 − 1 = 81. Thallium (Tl) is the element
with the atomic number 81, so we have that 207
207
0
81Tl → 82 Pb + −1e . 25. REASONING We can determine the identity of X in each of the decay processes by noting
that for each process, the sum of A and Z for the decay products must equal the values of A
and Z for the parent nuclei. SOLUTION
a. 211
82 Pb → 211 Bi + X
83 Using the reasoning discussed above, X must have A = 0 and Z = –1 . Therefore X must
be an electron, 0
–1 e; X represents a β – particle (electron) .
11
6 b. C → 11 B + X
5 Similar reasoning suggests that X must have A = 0 and Z = +1 . Therefore X must be a
positron; 0
+1 e; X represents a β + particle (positron) . 1600 NUCLEAR PHYSICS AND RADIOACTIVITY 231
90 c. Th * → 231 Th + X
90 Similar reasoning suggests that X must have A = 0 and Z = 0 . Therefore X must be a
gamma ray; X represents a γ ray .
210
84 d. Po → 206 Pb + X
82 Using the reasoning discussed above, X must have A = 4 and Z = 2 . Therefore X must be
a helium nucleus, 4 He ; X represents an α particle (helium nucleus) .
2 26. REASONING The γray photon is emitted when the nucleus changes from one energy state
to a lower energy state. The energy of the photon is the difference ∆E between the two
nuclear energy levels, in a way similar to that discussed in Section 30.3 for the energy levels
of the electron in the hydrogen atom. In that section, we saw that ∆E is related to the
frequency f and Planck’s constant h. Thus, we can determine the energy from a value for the
frequency. However, this value is not given. Instead, the wavelength λ is given. The
frequency can be obtained from the wavelength, since the two quantities are related to the
speed c of light. SOLUTION Section 30.3 discusses the fact that the photon emitted when the electron in a
hydrogen atom changes from a higher to a lower energy level has an energy ∆E, which is
the difference between the energy levels. A similar situation exists here when the nucleus
changes from a higher to a lower energy level. The γray photon that is emitted has an
energy ∆E given by
(30.4)
∆E = h f
The frequency is related to the wavelength according to
f= c (16.1) λ Substituting this expression into Equation 30.4 shows that ( c
∆ E = h f = h = 6.63 × 10−34 J ⋅ s
λ 3.00 ×10
) 1.14 ×10−11m/s = 1.74 ×10−14 J m 8 Finally, we convert this result from joules (J) to MeV: ( ) 1 eV 1 MeV ∆ E = 1.74 ×10−14 J = 0.109 MeV
−19 J 1×106 eV 1.60 ×10 Chapter 31 Problems 1601 27. REASONING Since energy is released during the decay, the combined mass of the lead
206
210
82 Pb daughter nucleus and the α particle is less than the mass of the polonium 84 Po
parent nucleus. The difference in mass is equivalent to the energy released. Since the recoil
of the lead nucleus is being ignored and we are assuming that all the released energy goes
into the kinetic energy KE of the α particle, it follows that released energy = KE = 1 mv 2
2 (Equation 6.2). Thus, the speed of the α particle is v = 2 ( KE )
.
m SOLUTION The decay reaction is
210
Po
1 84 3
4
24 209.982 848 u → 206
4
He
82 Pb3 + 1224
43
14
24 4.002 603 u 205.974 440 u The difference in the masses is
209.982 848 u − 205.974 440 u − 4.002 603 u = 5.805 × 10−3 u
This mass difference corresponds to an energy of (5.805 ×10−3 u ) 931.5uMeV = 5.407 MeV 1 This energy is the kinetic energy of the α particle (mass m = 6.6447 × 10–27 kg, see Example
2), so the speed of the α particle is v= 2 ( KE )
=
m 1.60 × 10−19 J 2 5.407 × 106 eV 1 eV = 1.61 × 107 m/s
6.6447 × 10−27 kg ( ) 28. REASONING When a nucleus undergoes an α decay, it emits one α particle that contains
two protons and two neutrons. Therefore, each α decay decreases the atomic mass number A
by four. A β − decay does not affect the atomic mass number A at all, so the difference
Ad − Ap between the atomic mass number Ad = 208 of the daughter nucleus atomic mass number Ap = 230 of the parent nucleus ( 208 Pb ) and the
82 ( 230 Rn ) after the emission of Nα alpha
86 particles is Ad − Ap = −4 Nα (1) 1602 NUCLEAR PHYSICS AND RADIOACTIVITY The negative sign in Equation (1) indicates a reduction in the atomic mass number in going
from the parent to the daughter. In order to determine the difference Zd − Zp between the
final atomic number Zd = 82 of the daughter nucleus and the atomic number Zp = 86 of the parent nucleus, we must take into account both the loss...
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