Physics 11: Subatomic Mass, Energy & Momentum
A.
Atomic Nucleus (30-1 to 30-8)
1. 2 types of particles—nucleons
a. positively charged proton and neutral neutron
b. rest mass chart (1 u = 1.66 x 10-27 kg)
Object
kg
u
Proton (11p)
1.67262 x 10-27
1.007276
Neutron (10n)
1.67493 x 10-27
1.008665
0
Electron ( -1e)
9.1094 x 10-31
0.00054858
2. nuclide (combination of protons and neutrons)
a. number of protons = atomic number (Z)
b. number of protons + neutrons = mass number (A)
c. nuclide symbol: AZX
1. X is atomic symbol with same Z number
2. A varies = isotopes
3. binding energy (BE) and nuclear forces
a. takes energy to break a stable nuclide into its parts
 energy is added to the nuclide to separate into
protons and neutrons
1. nuclide + BE = protons + neutrons
2. change in mass, mBE = BE/c2
a. c = 3 x 108 m/s (speed of light)
b. derived from E = mc2
c. mBE = mp + mn – mnuclide > 0
b. binding energy per nucleon (BE/# nucleons) is a
measure of nuclear stability
1. higher value = more stable
2. maximum around Fe-56
c. nuclear processes
1. exothermic when BE/#nucleon increases
2. mBE = mproducts – mreactants < 0
3. nuclear fusion: Zsmall  Zlarge
a. thermonuclear devise—Hydrogen bomb
b. sun's energy
4. nuclear fission: Zlarge  Zsmall
a. atomic bomb (Hiroshima)
b. nuclear power plant
4. strong nuclear force
a. acts over a very short distance (10-18 m)
b. binds "up" and "down" quarks together
1. proton = 2 up (u+2/3) + 1 down (d-1/3) quark
2. neutron = 1 up (u+2/3) + 2 down (d-1/3) quark
proton
neutron
binds neutrons and protons together
neutrons stabilize nuclide
1. electric repulsion between protons is significant
when protons > 10-18 m apart
2. neutrons increase total strong nuclear force
without increasing electric repulsion
naturally occurring radioactivity: unstable nuclides
undergo nuclear change
a. mass and charge number conserved
b. alpha emission (42)—low penetration
1. reduce total mass
2. 22688Ra  22286Rn + 42 (or 42He)
c. beta (0-1) & positron (01) emission—medium
1. down quark  up quark
2. d-1/3  u+2/3 +   10n  11p + 0-114 C  14 N + 0 (or 0 e)
6
7
-1
-1
3. u+2/3  d-1/3 + 11p  10n + 01
11 C  11 B + 0  (or 0 e)
6
5
1
1
d. gamma radiation, 00—high penetration
1. beta/positron collision E = 2mec2)
2. nuclear reactions (E = mBEc2)
c.
d.
5.
Name __________________________
e.
6.
7.
particles emitted with kinetic energy (part of m)
have a wave-like property (DeBroglie wavelength, )
1.  = h/p = h/mv
2. Planck's constant, h = 6.63 x 10-34 J•s
3. 23892U  23490Th + 42
mTh
m
234
4
momentum:
mThvTh = mv
kinetic energy: ½mThvTh2 << ½mv2
artificial induced radioactivity (transmutation)
a. nuclide bombarded by subatomic particle
b. nuclear reaction: 10n + 147N  146C + 11p
rate of decay and half-life
a. time it takes to reduce radioactivity by half is
constant = half life, t½
b. 1  1/2  1/4  1/8  1/16 ...
B.
Photons (27-2, 27-4, 27-4 to 27-7)
1. wave property
a. c = f—wavelength () and frequency (f)
b. electromagnetic spectrum
vibrating (hot) atoms and electron transitions in atoms
stopped
AC electric current
high-voltage
electrons
b.
graph of electron kinetic energy vs. frequency
1. kinetic energy measured as "stop voltage"
K
Kelectron = hf – 
slope = h
f
released
binding
energy
x-intercept = fthreshold

y-intercept = work function (ionization energy)

2.
C.
particle property: energy, E = hf (J)
a. color = energy, brightness = intensity
b. Compton effect: photon strikes free electron and
transfers some of its momentum to the electron
(photon wavelength increases)
1. p = h/(same as DeBroglie formula)
2. relativistic mass, m = p/c
3. photon formulas (kg, m, s, J)
c
E/p
In terms of:
f/
E = mc2
E = hf
Energy
E = pc
p = mc
Momentum
p = h/
Electron (27-1, 27-4, 27-8, 27-10 to 27-13)
1. electron properties
a. same charge as proton, but negative
b. mo = 9.11 x 10-31 kg (0.00054858 u)
2. electron structure in an atom: Bohr model
a.
3.
electrons occupy discrete energy levels
1. En = -B/n2 (BH = 13.6 eV)
2. n = energy level #
3. electron volt (eV): 1 eV = 1.6 x 10-19 J
b. change n by absorbing/emitting photons
1. Ephoton = En-high – En-low
2. EeV = 1240 eV•nm/nm
photoelectrons: Ephoton > Eionization
a. atom emits electron
# photons (intensity) =
# electrons (current)
4.
2. working version: KeV = 1240 eV•nm/nm - eV
electron formulas (9.11 x 10-31 kg, m, s, J)
v
E/p
In terms of:
f/
KJ = ½mv2
Energy
KJ = p2/2m
p = mv
Momentum
p = h/