Rohlf Chapter 11, #3,6,7,11,15,18,20,21,22,23,29,40

#3  Energy of incident photons and neutrons to give N a recoil energy of 1.4 MeV
a. Photon energy required comes from Compton scattering formula for scattering at 180°
At 180 degrees the wavelength change is given by 2hc/mc^2 where m is nitrogen mass
For the photon of wavelength w and energy E, delta w/w =delta E/E.
This shakes down to a photon energy E=sqrt(delta E mc^2/2)
The mass of the N-14 nucleus is 14.003*931.5 MeV/c^2 =    13043.7945 MeV/c^2
The photon energy required is then = sqrt(1.4 MeV*1.304E4 MeV/2) =    95.55 MeV/c^2
This is much more energy than you can get from a nuclear reaction, so the particle is not a photon.

b. Neutron energy required comes from elastic collision.
For m1<<m2 and headon, the change in momentum for m1 is twice its original momentum
(I.e., it bounces straight back with essentially its original speed.)
.5*mN*vN^2 = 1.4 MeV = E, but mN*vN=2m(neutron)*v(neutron),
so KE neutron = Em(nitrogen)/(4*m(neutron)) = (1.4 MeV)(1.304E4 MeV)/(4*940MeV)
KE neutron = 4.855 MeV

#6 Mass of U-235 nucleus and atom
92 protons at 938.2723 MeV =  86321051.60000 MeV/c^2
143 protons at 939.5656 MeV=  134357880.80000 MeV/c^2
minus binding energy 235*7.59=  1783.65000 MeV/c^2
Net mass of nucleus =  220677148.75000 MeV/c^2
Add electrons
 =92*.511 MeV =  47.01200 MeV/c^2
Mass of atom =  220677195.76200 MeV/c^2
#7 Binding energies for manganese, iron and cobalt
For Mn-55, E/A=482.1/55=  8.765454545 MeV
For Fe-54, E/A=471.8/54=  8.737037037 MeV
For Fe-56, E/A=492.3/56=  8.791071429 MeV
For Fe-57, E/A=499.9/57=  8.770175439 MeV
For Fe-58, E/A=510.0/58=  8.793103448 MeV
For C0-59, E/A=517.3/59=  8.76779661 MeV

#11. The binding energy for Fe-55 is 481 MeV.
The Weizsaecker formula gives
E(odd-even)=15.75*55-17.8*55^(2/3)-.711*26^2/55^(1/3)-23.7*9/55 =
    = 478.55 MeV

For Co-57 the binding energy is 498 MeV
E(odd-odd)=15.75*57-17.8*57^(2/3)-.711*27^2/57^(1/3)-23.7*9/57-11.18/sqrt(57) =
    = 494.21 MeV

For Ni-58 the binding energy is 506 MeV. The Weizsaecker formula gives
E(even-even)=15.75*58-17.8*58^(2/3)-.711*28^2/58^(1/3)-23.7*4/58+11.18/sqrt(58) =
   = 502.62 MeV

#15 A factor of 32 is 5 halflives, halflife=1 s

#18 Age = half-life*ln(R0/R)/ln(2)=5730y*ln(25/16.7)/ln(2)=    3335.3 years

#20 Neutron binding energies.
a. Last neutron in O-17: Energy = 131.77-127.62=   4.15 MeV
b. Last neutron in O-18:Energy =139.81-131.77=   8.04 MeV

#21Fe-55 + electron=Mn-55+neutrino
b. The xray is likely a K-alpha xray for Mn, so 13.6*(25-1)^2*3/4 using Z-1 as effective nuc charge
So a reasonable guess is =  5875.2 eV

#22 Identify the unknown
a. X has Z=13 and A=23 which is Al-23
b. X has Z=78 and A=190 which is Pt-190
c. X has Z=5 and A=11 which is B-11
d. Z has decreased, so positron
e. X has Z=1 and A=1 which is a proton

#23  Energy release = mass of alpha + U235-Pu239 =
    =28.296+1783.897-1806.948 =   5.245 MeV

KE of alpha=Q(A-4)/A=5.25*235/239=   5.162 MeV

#29 Source speeds for given energy change
a. v= c*delta E/E =3E8m/s*1E-5/129000 =   0.0233 m/s
b. v= c*delta E/E =3E8m/s*1E-8/14400 =   0.000208 m/s

#40 Since one gram corresponds to 1 Curie, then 0.1 microcurie
1e-10kg*6E23/(226*1E-3)=  2.65487E+14 atoms