Test 1 Flashcards
Kilo- (k)
1000 units (10^3)
Hecto- (h)
100 units
Deka- (da)
10 units
Deci- (d)
0.1 units
Centi- (c)
0.01 units (10^-2)
Milli- (m)
0.001 units (10^-3)
Mega- (M)
1,000,000 units (10^6)
Speed of light
3 x 10^8 m/s
3 subatomic particles
Neutrons
Electrons
Protons
Neutrons and protons inside the nucleus
Nucleons
2 nucleons
Neutrons
Protons
Radius of the nucleus
10^-15 m
Radius of the electronic orbit of electrons around the nucleus
10^-10 m
The nucleus orbit is _______ than the electron orbit
Smaller
The mass of a nucleon is about _______ times that of an electron
2,000
A theory of atomic structure in which an atom is assumed to consist of protons as nucleons in the nucleus, with electrons moving in distinct circular orbits around it, each orbit corresponding to a specific quantized energy state
Bohr’s model
Max number of electrons in each respective shell
2n^2
Electrons closer to the nucleus have ________ binding energy
Higher
Number of protons
Atomic number (Z)
Number of nucleons
Atomic mass number (A) (amus)
Formula for the number of neutrons
N=A-Z
How is the chemical identity of an element determined?
By the number of protons in the nucleus
What determines an element’s chemical behavior?
The number of electrons
Two atomic nuclei with the same atomic number/Z/number of protons but a different number of neutrons
Isotope
Same number of neutrons but different atomic number (Z)
Isotone
Same mass number/number of nucleons (protons + neutrons)/A, but different atomic number/number of protons/Z
Isobar
Same mass number/number of nucleons/A, but in a different nuclear state (metastable state/different energy level = excited state)
Isomer
Every gram atomic weight of a substance contains ______ number of atoms
The same
Avagadro’s number (NA)
6.0221 x 10^23 atoms per gram atomic weight (mole) or electrons/gram
Atomic mass unit
amu
1 amu = how many kg or Mev
- 66 x 10^-27
931. 4 MeV
Formula to find the number of atoms per gram for an element
Avagadro’s number (NA)/atomic weight (AW)
6.0221 x 10^23 atoms/gram / AW
1 amu is equal to what of a Carbon-12 atom?
1/12 of a Carbon-12 atom
When subatomic particles join together to form an atom it takes energy to do so; the subatomic particles give up some of their mass to be converted to attain this necessary energy to hold the particles together
Difference of the mass of an element versus the mass of all subatomic particles in that particular atom
Mass defect
Amount of energy required to remove an electron from the atom
Binding energy
Mass of a proton
1.00727 amu
Mass of a neutron
1.00866 amu
Mass of an electron
0.000548 amu
Formula for mass defect
Atomic mass number - ((# of P+ * 1.00727) + (# of N * 1.00866) + (# of E- * (0.000548))
Einstein’s Theory of Relativity
Energy (E) = mass (m) * speed of light (C)^2
kgm^2/s^2
Joules (J)
1 eV = how many J?
1.60218 x 10^-19 J
5 steps to find binding energy
Find the mass defect (amu) Convert it to kilograms (kg) Find the energy converted (E=mc^2) Convert to eV Convert to megaelectronvolts (MeV)
Formula for converting mass defect (amu) to kg
Mass defect (amu) x (1.66 x 10^-27 kg/1 amu) = kg
Formula for converting energy (J) to eV
eV = J/1.602x10^-19 eV
Formula for converting eV to MeV
MeV = eV/1,000,000
Basic unit of energy
Joule (J)
1 J/kg = ? Rads = ? Gy
1 J/kg = 100 Rads = 1 Gy
100 cGy = ? Gy
100 cGy = 1 Gy
1 cGy = ? rad = ? Gy
1 cGy = 1 rad = 0.01 Gy
Combination of two lighter nuclei that takes energy to put them together; low mass nuclei are combined to produce a larger nucleus
Nuclear reaction in which atomic nuclei of low atomic number fuse to form a heavier nucleus with the release of energy
If light energy could combine, the average binding energy of the resulting nucleus would be greater, leaving excess energy to be released
Occurs in nature
Fuse two small particles to make a big one
Nuclear fusion
Nucleus with an atomic number greater than 56 splits into two smaller nuclei and have a higher binding energy per nucleon and therefore energy is released (ex: atomic bomb or Uranium Nuclear Reactors split atoms to give off energy)
Occurs when high Z nuclei are bombarded by neutrons; after absorbing the neutrons, it splits into nuclei of lower Z, as well as more neutrons
Ex: (235/92)U + (1/0)n –> (236/92)U –> (141/56)Ba + (3)(1/0)n + Q (energy)
Nuclear fission
As nature attempts to balance forces, spontaneous transformation of a nucleus into a lower binding energy occurs
This larger nucleus breaks into two or more parts that can be radioactive themselves (alpha particles, beta, etc.); excess energy is released as gamma rays and a new product called the daughter is more tightly bound (higher binding energy)
Nature attempts to minimize energy/make it as negative as possible by transforming one nucleus into another with lower (more negative) binding energy; this excess energy is released as radiation
Nuclei are breaking apart to become stable
Radioactive decay
Resulting nucleus of radioactive decay that is more tightly bound
Some radioactive substances break down and give rise to a radioactive product
Daughter
Nuclei that do not undergo radioactive decay
Stable
Wave model (energy)
C=vλ
C=velocity
v=frequency (Hz or 1/sec)
λ=wavelength
Describes the relationship between energy and frequency (λ)
Plank’s constant
Graphs binding energy per nucleon vs. atomic number
Curve of binding energy (BE)
Average binding energy (BE) of most nuclei
8 MeV per nucleon
BE per nucleon reaches peak with what element?
Iron (Fe56)
Excess energy is released from radioactive decay as this
Gamma rays
Too many protons make the nucleus _______
Unstable
Ratio of neutrons to protons
1.4 neutrons for 1 proton
A material composed of the antiparticle “partners” to the corresponding particles of ordinary matter
A particle and its antiparticle have the same mass as one another, but opposite electric charge and other quantum numbers
Antimatter
Particle with equal mass and magnitude to an electron but opposite sign of charge (+)
Postiron (e+)
Anti-electron
Every particle has an ______
Antiparticle
When positron meets electron, they disappear, leaving behind to gamma ray photons that travel in opposite directions
This is an example of the complete conservation of matter into energy as described by Einstein’s equation E=mc^2
Charge is conserved because the net charge both before and after is zero
Annihilation reaction
e+ +e- = 2y
What is the energy of each gamma ray emitted during annihilation reaction?
0.511 MeV
Gamma radiation emitted during annihilation reaction
Annihilation radiation
The total energy of the two gamma photons emitted during annihilation reaction is equal to what?
The rest mass energy of the positron plus electron