Sustainable Energy 1st Edition By Richard Dunlap - Test Bank

Sustainable Energy 1st Edition By Richard Dunlap – Test Bank

 

Instant Download – Complete Test Bank With Answers

 

 

Sample Questions Are Posted Below

 

Chapter 5

Some Basic Nuclear Physics

 

5.1 ⁸Be is one of the few light nuclei that decays by α–decay and it decays by splitting into two α-particles (⁴He nuclei). Using the following neutral atom masses calculate the total energy released during this process: m(⁸Be) = 8.00530594 u and m(⁴He) = 4.00260325 u.

 

Solution The decay process is

 

⁸Be → ⁴He + ⁴ He

 

Using atomic masses as given above properly accounts for all electron masses. Thus

 

E = [m(⁸Be) – 2 × m(4He)

                                                                     = (8.00530594 u – 2 × 4.00260325 u) × 931.494 MeV/u

= 0.093 MeV

 

 

5.2 The mass of a neutral ⁴⁰Ca atom is 39.96259116 u. Calculate the total binding energy of this atom.

 

Solution The binding energy is

 

 

Ca has an atomic number of Z = 20, so a neutral ⁴⁰Ca atom has 20 protons, 20 neutrons and 20 electrons. Substituting these values and the masses of the particles from Appendix II gives the binding energy as

 

B = (20 × 1.008664904 u + 20 × (1.007276470 +

0.0005485799) u – 39.96259116 u) × 931.494 MeV/u

= 342.1 MeV

 

 

5.3 A sample of ²²Na (half-life = 2.6 years) decays at a rate of 10¹² decays per second.  How long will it take the decay rate to decrease to 100 decays per second?

 

Solution The radiation decay law gives

 

 

Then the decay rate is  or

 

 

Then at t = 0

 

 

so

 

 

or

 

 

Calling the left hand side R then the above is written as

 

 

and solving for t gives

 

,

 

where R is the relative decay rate at time t. In this problem the relative decay rate is 10^–10, i.e. [10²/10¹²]. The decay constant is given by

 

 

then

 

 

 

5.4 What are the decay products of the β⁺ decay of the following nuclides: ¹¹C, ²³Mg, ⁷⁷Br and ¹³⁶La?

 

Solution The β⁺ decay process leaves A unchanged and decreases Z by 1 and increases N by 1 so for the following nuclides

 

 

 

 

nuclide	A	N	Z
11C	11	5	6
23Mg	23	11	12
77Br	77	42	35
136La	136	79	57

 

Thus the progeny are determined to be

 

A	N	Z	progeny
11	6	5	11B
23	12	11	23Na
77	43	34	77Se
136	80	56	136Ba

 

 

5.5 Tabulate the number of protons, electrons, and neutrons, for neutral atoms of the following β-stable nuclides; ⁵²Cr, ⁹⁰Zr, ¹²⁰Sn, ¹⁵⁶Gd and ²⁴¹Am and show that for these nuclides the ratio of neutrons to protons increases with increasing nuclear mass.

 

Solution The superscript gives A = N + Z. The value of Z is found in the periodic table for these elements to be

 

Z(Cr) = 24

Z(Zr) = 40

Z(Sn) = 50

Z(Gd) = 64

Z(Am) = 95

 

Since Z gives both the number of electrons and protons for the neutral atom, and N = A − Z then we find the following numbers of protons, electrons and neutrons and the ratio of N/Z.

 

nuclide	protons (Z)	electrons	neutrons (N)	N/Z
52Cr	24	24	28	1.17
90Zr	40	40	50	1.25
120Sn	50	50	70	1.40
156Gd	64	64	92	1.44
241Am	95	95	146	1.54

 

 

5.6 Calculate the energy released during the α–decay of ²³⁶U. The following atomic masses may be of use: m(²³⁶U) = 236.0455619 u, m(²³²U) = 232.0371463 u, m(²³⁴Th) = 234.0435955 u, m(²³²Th) = 232.0380504 u and m(⁴He) = 4.00260325 u.

 

Solution The decay process is

 

²³⁶U → ²³²Th + ⁴He

 

Using atomic mass properly accounts for all electrons with the masses given above and

 

E = [m(²³⁶U) – m(²³²Th) – m(⁴He)]c²

                                                                = (236.0455619 u – 232.0380504 u – 4.00260325 u)c²

= (4.908 × 10^–3 u) × 931.494 MeV/u

= 4.57 MeV