#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