Problem#1
(a) Show that the expression for the Planck length, √(ћg/c3) has dimensions of length. (b) Evaluate the numerical value of √(ћg/c3).
Answer:
(a) The dimensions of √(ћg/c3) are
[(E.T)(E.L/M2)/(L.T-1)3]1/2 = (E/M)(T2/L) = L
(b) the numerical value of √(ћg/c3) = [(6.626 x 10-34 Js)(6.673 x 10-11 N.m2/kg2)/(2π x 3.00 x 108 m/s)3]1/2
= 1.62 x 10-35 m.
Problem#2
Calculate the energy released in each reaction: (a) p + 2H → 3He, (b) n + 3He → 4He.
Answer:
(a) p + 2H → 3He can write 1H + 2H → 3He
If neutral atom masses are used then the masses of the two atomic electrons on each side of the reaction will cancel.
the mass decrease is
∆m = m(1H) + m(2H) – m(3He)
∆m = 1.007825 u + 2.014102 u – 3.016029 u = 0.005898 u
The energy released is the energy equivalent of this mass decrease:
E = ∆mc2
E = (0.005898 u)c2 = 0.005898(931.5 MeV/c2) = 5.494 MeV
(b) n + 3He → 4He.
If neutral helium masses are used then the masses of the two atomic electrons on each side of the reaction equation will cancel. The mass decrease is
∆m = m(n) + m(3He) – m(4He)
∆m = 1.008665 u + 3.016029 u – 4.002603 u = 0.022091 u
The energy released is the energy equivalent of this mass decrease:
E = ∆mc2
E = (0.022091u)c2 = 0.022091(931.5 MeV/c2) = 20.58 MeV
Problem#3
Calculate the energy (in MeV) released in the triple-alpha process 34He → 12C.
Answer:
The energy released in the reaction is the energy equivalent of the mass decrease that occurs in the reaction.
1u is equivalent to 931.5 MeV, so that
E = ∆mc2
E = [3m(4He) – m(12C)]c2
E = [3(4.002603 u) – 12.00000u]c2
E = 7.27 MeV
Problem#5
Calculate the reaction energy Q (in MeV) for the reaction e–1 + p → n + ve. Is this reaction endoergic or exoergic?
Answer:
When Q is negative the reaction is endoergic. When Q is positive the reaction is exoergic.
The mass decrease is
∆m = me + mp – mn – mve
∆m = 0.0005486 u + 1.007276 u - 1.008665 u = –0.00084u
The energy released is the energy equivalent of this mass decrease:
E = ∆mc2
E = (–0.00084)(931.5 MeV/c2) = –0.783 MeV and is endoergic.
Problem#6
Calculate the reaction energy Q (in MeV) for the nucleosynthesis reaction
12C6 + 4He2 → 16O8
Is this reaction endoergic or exoergic?
Answer:
The mass decrease is
∆m = m(12C6) + m(4He2) – m(16O8)
∆m = 12.00000u + 4.002603u – 15.994915u = 0.00769u
The energy released is the energy equivalent of this mass decrease:
E = ∆mc2
E = (0.00769)(931.5 MeV/c2) = 7.16 MeV an exoergic reaction.
Problem#7
The 2.728-K blackbody radiation has its peak wavelength at 1.062 mm. What was the peak wavelength at t = 700,000 y when the temperature was 3000 K?
Answer:
The Wien displacement law
λmT= constant
λm,1T1 = λm,2T2
λm,1(2.728-K) = (1.062mm)(3000K)
λm,1 = 966mm
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