Physics Review 6
A.
B.
Atomic Nucleus
1. nuclide symbol: AZX
2. binding energy (BE)
a. mBE = BE/c2 (c = 3 x 108 m/s)
b. mBE = mp + mn – mnuclide > 0 (nonspontaneous)
c. mBE = mproducts – mreactants < 0 (spontaneous)
3. naturally occurring radioactivity
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. d-1/3  u+2/3 +   10n  11p + 0-114 C  14 N + 0 (or 0 e)
6
7
-1
-1
2. 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)
e. particles emitted with kinetic energy (part of m)
have a wave-like property (DeBroglie wavelength, )
1.  = h/p = h/mv (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
4. transmutation: 10n + 147N  146C + 11p
5. half-life (t½): 1  1/2  1/4  1/8  1/16 ...
Photons
1. c = f—wavelength () and frequency (f)
electromagnetic spectrum
Name __________________________
Practice Free Response
1.
232
92U (232.037146 u) undergoes alpha (4.002602 u) decay
to produce an isotope of Th (228.028731).
a. Write a nuclear equation.
b.
Calculate the change in mass in u.
c.
Calculate the change in mass in kg (1 u = 1.66 x 10-27 kg)

2.
d.
Calculate the energy in joules released.
e.
Assume the alpha particle absorbs all the energy in the
form of kinetic energy. What is the speed of the alpha
particle (ignore relativistic effects)?
f.
What is the momentum of the alpha particle?
The diagram shows the lowest four energy levels of an atom.
An electron makes the transition from n = 4 to n = 1 and
emits a photon.
-5.00 eV
-11.25 eV
-45.00 eV
Determine
a. the electron energy for n = 4.

2.
C.
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
1. Bohr model
a. 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
2. photoelectrons
a. formula: Kelectron = Ephoton – 
working version: KeV = 1240 eV•nm/nm - eV
3. electron formulas (9.11 x 10-31 kg, m, s, J)
v
E/p
In terms of:
f/
K = ½mv2
Energy
K = p2/2m
p = mv
Momentum
p = h/
b.
the energy of the emitted photon in eV.
c.
the wavelength of the emitted photon.
d.
the momentum of the emitted photon.
The photon is incident on silver, which emits a photoelectron
(m = 9.11 x 10-31 kg) with wavelength  = 4.90 x 10-10 m.
e. What is the momentum of the photoelectron in kg•m/s?
f.
What is the kinetic energy of the photoelectron in eV?
g.
What is the work function of silver?