Unformatted text preview: of two protons for each α particle
emitted, and the gain of one proton for each β − particle emitted. Therefore, we have that Zd − Z p = −2 Nα + N β (2) where Nβ is the number of β − particles emitted in the decay series.
SOLUTION Solving Equation (1) for Nα, we obtain
Nα = Ad − Ap
−4 = 208 − 220
=3
−4 Solving Equation (2) for Nβ yields N β = 2 Nα + Z d − Z p = 2(3) + 82 − 86 = 2
Therefore, in this decay series, 3 α particles and 2 β − particles are emitted. 29. REASONING AND SOLUTION The β − decay reaction is
208
81 Tl → 208 Pb + −0 e
82
1 so
A
Z X → 208 Pb + 4 He
82
2 gives
A
Z X = 212 Po
84 30. REASONING As Section 7.2 discusses, the principle of conservation of linear momentum
indicates that the total momentum of an isolated system is conserved. Assuming that no
external net force acts on the beryllium nucleus, it is isolated and the emission of the gamma
ray satisfies this principle. The magnitude of the momentum of the gamma ray photon is
h/λ, according to Equation 29.6, where h is Planck’s constant and λ is the wavelength. We
take the direction of the photon as the positive direction. Then, using Equation 7.2, we write
the momentum of the recoiling beryllium nucleus as –mvBe, where m is the mass and vBe is Chapter 31 Problems 1603 the speed. The conservation principle indicates that the total momentum after the emission
must be equal to the total momentum before the emission, so we have h
− mv Be
+
=
0
1 24
43
14243
λ
14243 Total momentum
Momentum of nucleus
Momentum of gamma
ray photon after emission before emission This expression can be solved for the wavelength.
SOLUTION Solving the conservationofmomentum expression, we obtain λ= h
mv Be Before we can use this equation, the mass of the beryllium must be converted from atomic
mass units to kilograms: b m Be = 7 .017 u F
g1.6605 × 10
G 1u
H −27 kg I = 1.165 × 10
J
K −26 kg The wavelength, then, is λ= h
6.63 × 10 −34 J ⋅ s
=
= 2 .60 × 10 −12 m
−26
4
mv Be
1.165 × 10
kg 2 .19 × 10 m / s c h
c h 31. SSM REASONING Energy is released during the β decay. To find the energy released,
we determine how much the mass has decreased because of the decay and then calculate the
equivalent energy. The reaction and masses are shown below:
22
Na
11
13
2 21. 994 434 u → 22
Ne
10
13
2 21.991 383 u + 0
+1 ;e 5.485 799 × 10 –4 u SOLUTION The decrease in mass is c h 21.994 434 u – 21.991 383 u + 5.485 799 × 10 –4 u + 5.485 799 × 10 –4 u = 0.001 954 u
where the extra electron mass takes into account the fact that the atomic mass for sodium
includes the mass of 11 electrons, whereas the atomic mass for neon includes the mass of
only 10 electrons.
Since 1 u is equivalent to 931.5 MeV, the released energy is 1604 NUCLEAR PHYSICS AND RADIOACTIVITY F
0
b.001 954 u g931.5uMeV I =
G1 J
H
K 1.82 MeV 32. REASONING The basis of our solution is the fact that only onehalf of the radioactive
nuclei present initially remain after a time equal to one halflife. After a time period that
11
1
or
×
22
4
111
1
or of the
××
222
8 equals two halflives, the number of nuclei remaining is of the initial number. After three halflives, the number remaining is initial number, and so on. Thus, to determine the fraction of nuclei remaining after a given time period, we need
only to know how many halflives that period contains.
SOLUTION For sample A, the number of nuclei remaining is 111
=×
422 of the initial number. We can see, then, that the time period involved is equal to two half lives of
1
2 radioactive isotope A. But we know that T1/2, B = T1/2, A , so that this time period must be
equal to four halflives of radioactive isotope B. The fraction f of the B nuclei that remain,
therefore, is
f= 1111
×××
2222 = 1
16 33. SSM REASONING AND SOLUTION The number of radioactive nuclei that remains in a
sample after a time t is given by Equation 31.5, N = N 0 e – λ t , where λ is the decay
constant. From Equation 31.6, we know that the decay constant is related to the halflife by
T1/ 2 = 0 .693 / λ ; therefore, λ = 0 . 693 / T1/ 2 and we can write
N
= e –( 0 . 693/ T1/2 ) t
N0 or t
1
=–
ln
T1/2
0.693 FN I
GJ
N
HK
0 When the number of radioactive nuclei decreases to onemillionth of the initial number,
N / N 0 = 1.00 × 10 –6 ; therefore, the number of halflives is t
1
=–
ln (1.00 × 10 –6 ) = 19.9
T1/2
0.693 Chapter 31 Problems 1605 34. REASONING The halflife T1/2 is inversely related to the decay constant λ by the relation T1 / 2 = 0.693 / λ (Equation 31.6). To convert from seconds to days, we use the fact that
3600 s = 1 h and 24 h = 1 day. SOLUTION The halflife of radium
T1/ 2 = 0.693 λ = 224
88 Ra is 0.693
= 3.66 days
−1 ) 3600 s 24 h s 1 h 1 day ( 2.19 ×10−6 35. REASONING AND SOLUTION The amount remaining is 0.0100% = 0.000 100. We know N / N0 = e –0.693t /T1/2 . Therefore, we find
t=− N
29.1 yr
ln ln(0.000 100) = 387 yr
=− 0.693 N0 0.693 T1 / 2 36. REASONING The number N of radioactive nuclei remaining after a time t is given by
N =...
View
Full Document
 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

Click to edit the document details