6.272. T > Eb (MP nic1)1Md = 3.3 MeV.
6.273. Between 1.89 and 2.06 MeV.
6.274. Q = — 1142 To t = —3.7 MeV.
6.275. 1.88 and 5.75 MeV respectively.
6.276. 4.4 MeV; 5.3 . 106 m/s.
6.277. T.= ..± 4 [(m4 — mt)T mm i+ 2m7n 4 a T,h ]=2.2
mi, m2, m3, my are the masses of neutron, a Ciz
alpha particle, and a Be nucleus.
6.278. By Eb/2mc2 = 0.06%, where in is the mass of
6.279. E = Q
2/3T = 6.5 MeV.
6.280. E1 = Eb+ md7mc T, 16.7, 16.9, 17.5 and
where Eb is the binding energy of a deuteron in the
nucleus.
6.281. a = (M/Np d) in 1 = 2.5 kb, where M is the molar mass
of cadmium, N is the Avogadro number, p is the density of cadmium.
6.282. T oll. = exp [(2a1 ± a 2) nd] = 20, where n is the concentration of heavy water molecules.
6.283. w = (1 — exp [ — (as + aa) nd]) asl(as-F-cra)= 0.35,
where n is the concentration of Fe nuclei.
6.284. (a) T = (w1k) In 2; (b) w = ATe/It In 2 =
6.285. (a) t =
= 3.106 years; (b) N max = Jallo T/ln 2 =
= 1.0.1013, where No is the number of Atli" nuclei in the foil.
6.286. N = (1 — e-lt) Jna/X.
6.287. J = AemIaN 0 (1 — e-lt) = 6.106 part./(cm2 .$), where X
is the decay constant, No is the number of Au nuclei in the foil.
6.288. N = Noki-1 = 1.3.106, where i is the number of generations.
6.289. N = vP/E = 0.8.1016 s -1.
6.290. (a) N/No = 4.102; (b) T = ti/(k — 1) = 10 s.
6.291. 0.05, 0.4, and 9 GeV respectively.
6.292. (/)=c-ro
(II+ 2)=15 m.
6.293. To = lmcIV T (T
2mc2) = 26 ns, where in is the rest
mass of a pion.
6.294. J/J0 =exp — /mc/To V T (T -1-2mc2)1= 0.22, where m is
the rest mass of a negative pion.
6.295*. Tp. = (ma — mµ)2/2m. = 4.1 MeV, Ev = 29.8 MeV.
6.296*. T = [(mZ — m„)2 — mMI2mE= 19.5 MeV.
6.297*. Tmax = (Mu
me)2 /2mi, = 52.5 MeV.
6.298*. m= mp + T
T (T 2mp)=1115 MeV, a A
particle.
6.299*.
= 112 (ml — 714)1(mn T) = 22 MeV.
MeV, where
nucleus, an
a deuteron.
17.7 MeV,
transitional
* In the answers to Problems 6.295- 6.299 marked ',by an asterisk the
quantity mc2 is abbreviated as m.
6.273. Between 1.89 and 2.06 MeV.
6.274. Q = — 1142 To t = —3.7 MeV.
6.275. 1.88 and 5.75 MeV respectively.
6.276. 4.4 MeV; 5.3 . 106 m/s.
6.277. T.= ..± 4 [(m4 — mt)T mm i+ 2m7n 4 a T,h ]=2.2
mi, m2, m3, my are the masses of neutron, a Ciz
alpha particle, and a Be nucleus.
6.278. By Eb/2mc2 = 0.06%, where in is the mass of
6.279. E = Q
2/3T = 6.5 MeV.
6.280. E1 = Eb+ md7mc T, 16.7, 16.9, 17.5 and
where Eb is the binding energy of a deuteron in the
nucleus.
6.281. a = (M/Np d) in 1 = 2.5 kb, where M is the molar mass
of cadmium, N is the Avogadro number, p is the density of cadmium.
6.282. T oll. = exp [(2a1 ± a 2) nd] = 20, where n is the concentration of heavy water molecules.
6.283. w = (1 — exp [ — (as + aa) nd]) asl(as-F-cra)= 0.35,
where n is the concentration of Fe nuclei.
6.284. (a) T = (w1k) In 2; (b) w = ATe/It In 2 =
6.285. (a) t =
= 3.106 years; (b) N max = Jallo T/ln 2 =
= 1.0.1013, where No is the number of Atli" nuclei in the foil.
6.286. N = (1 — e-lt) Jna/X.
6.287. J = AemIaN 0 (1 — e-lt) = 6.106 part./(cm2 .$), where X
is the decay constant, No is the number of Au nuclei in the foil.
6.288. N = Noki-1 = 1.3.106, where i is the number of generations.
6.289. N = vP/E = 0.8.1016 s -1.
6.290. (a) N/No = 4.102; (b) T = ti/(k — 1) = 10 s.
6.291. 0.05, 0.4, and 9 GeV respectively.
6.292. (/)=c-ro
(II+ 2)=15 m.
6.293. To = lmcIV T (T
2mc2) = 26 ns, where in is the rest
mass of a pion.
6.294. J/J0 =exp — /mc/To V T (T -1-2mc2)1= 0.22, where m is
the rest mass of a negative pion.
6.295*. Tp. = (ma — mµ)2/2m. = 4.1 MeV, Ev = 29.8 MeV.
6.296*. T = [(mZ — m„)2 — mMI2mE= 19.5 MeV.
6.297*. Tmax = (Mu
me)2 /2mi, = 52.5 MeV.
6.298*. m= mp + T
T (T 2mp)=1115 MeV, a A
particle.
6.299*.
= 112 (ml — 714)1(mn T) = 22 MeV.
MeV, where
nucleus, an
a deuteron.
17.7 MeV,
transitional
* In the answers to Problems 6.295- 6.299 marked ',by an asterisk the
quantity mc2 is abbreviated as m.