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• (21) Hydrogen Atom spectrum levels, …
• C is false Method
• Background: ** Skip to the next problem if you are in a hurry.  (28) Counting Statistics
• Why are the Pauli matrices introduced
• If you are not familiar with modern physics, skip this paragraph.
• Special properties of the Pauli matrices
• Actions of the Pauli Set
• (28) Counting Statistics
• (29) Graphs and Basic Lab
• (30) Lagrangian and Hamiltonian Mechanics
• Conservation or constants of the motion
• Kinetic energy in terms of the center of mass and relative coordinates
• There are three fundamental processes in beta decay

Transition metals  filling d sub-shell*

24	Cr	(Ar)(4s)2(3d)4	7S3
25	Mn	(Ar)(4s)2(3d)5	6S5/2
26	Fe	(Ar)(4s)2(3d)6	5D4
27	Co	(Ar)(4s)2(3d)7	4F9/2
28	Ni	(Ar)(4s)2(3d)8	2F4
29	Cu	(Ar)(4s)2(3d)9	2S½
30	Zn	(Ar)(4s)2(3d)10	1S0	Sub-shell filled  spherical, zero angular momentum 1S0
31	Ga	(Ar)(4s)2(3d)10(4p)1	2P½	Filling the 2p sub-shell
32	Ge	(Ar)(4s)2(3d)10(4p)2	3P0	Maximize unpaired electrons to minimize exchange energy
33	AS	(Ar)(4s)2(3d)10(4p)3	4S3/2	Half filled sub-shell; max net spin J = |L – S|
34	Se	(Ar)(4s)2(3d)10(4p)4	3P2	Spin decreases as electrons added pairing previous ones
35	Br	(Ar)(4s)2(3d)10(4p)5	2P3/2	J = |L + S|
36	Kr	(Ar)(4s)2(3d)10(4p)6	1S0	Sub-shell filled  spherical, zero angular momentum 1S0

The fine structure energies complicate the filling sequence post Kr so we stop. *The sequences of elements arising while an inner f sub-shell is filling are called rare earths.

Electron shell filling sequence. Except for (H, He), the first electron in an S shell leads to an alkali metal (strong electron donor). Two s electrons yield a metal. The nth valence shell consists of the two ns and six np states. Beginning with period n = 4, inner shell (d and f electrons of lower n) fill after the nS, but before the nP. Shells fill first with spins parallel (say up) and add spins anti-parallel only after the sub-shell is half filled. Elements with approximately half-filled d and f sub-shells exhibit magnetic properties associated with the unpaired spins. The D, F, shells fill inside the valence electrons and have little impact on bounding. The transition metals and rare earths are the large group of elements differing by the number of electrons in the partially filled D and F sub-shells.

Ex 20.) For which of the following elements is the ionization energy of a neutral atom the lowest? (Z) is the atomic number.

(A) Oxygen (Z = 8) (B) Fluorine (Z = 9) (C) Neon (Z = 10)

(D) Sodium (Z = 11) (E) Magnesium (Z = 12)
The first period of elements includes hydrogen and helium with one and two electrons. In the next period (3 , there are up to eight electrons in the valance shell. Oxygen and fluorine need electrons to close the shell and so are hard to ionize. Neon is the very stable closed shell configuration. Sodium has just one electron outside a closed shell and so is easy to ionize. For Magnesium, the nucleus plus inner electron shells combined have an effective charge of plus 2 and so the screened nuclear charge is larger and it binds the outer electrons more tightly than does the sodium nucleus plus inner shell electrons (net charge +1). D
Sodium is a light alkali metal and so is an electron donor. In compounds such as NaCl, it is strongly ionic with the sodium being positive which means its electron has transferred to the chlorine. After hydrogen, the Group I atoms are willing electron donors and easy to ionize. Group 7 is the group of the strongest electron acceptors. Top Left  strong donor; Bottom Right (grp.7)  strong acceptor.
(21) Hydrogen Atom spectrum levels, …


, a_o = 0.0529 nm for hydrogen,  reduced mass, Z nuclear charge

The radius varies as (the mass of the orbiting particle)^-1 and as (the nuclear charge)^-1.

H-atom Transition Table

Lyman Series (n  1) Series Limit: n_upper =  to n_series

Transition						Series Limit
Designation	L	L	L	L	L	L: 13.6 eV
Wavelength	 nm	102 nm	97.2 nm	94.9 nm	93.8 nm	91.2 nm
Color	VUV	VUV	VUV	VUV	VUV	VUV

Balmer Series (n  2)

Transition						Series Limit
Designation	H	H	H	H	H	H: 3.4 eV
Wavelength	 nm	486.1 nm	434.1 nm	410.2 nm	397.0 nm	364.6 nm
Color	red	blue-green	violet	violet	violet	UV

Paschen Series (n  3)

Transition						Series Limit
Designation	P	P	P	P	B	P: 1.51 eV
Wavelength	1876 nm	1282 nm	1094 nm	1005 nm	955 nm	820.6 nm
Color	IR	IR	IR	IR	IR	near IR

Brackett (4) limit 1458.03 nm; Pfund (5) limit 2278.17 nm; Humphreys (6) 3280.56 nm
Ex 21.) The Paschen series for hydrogen corresponds to transitions that end in states with quantum number n = 3. The shortest wavelength line in the Paschen series is closest to which of the following? (The ionization energy of hydrogen is 13.6 eV and hc = 1240 eV nm)

(A) 125 nm (B) 250 nm (C) 400 nm (D) 800 nm (E) 1800 nm

The shortest wavelength would correspond to the transition with the largest E from the highest bound level down to the n = 3 level. You are given 13.6 eV; You must deduce E_n = -13.6 n^-2.

;

(22) Blackbody radiation

Total power per area = Stephan Boltzmann constant times T to the fourth power =  T ⁴

Wien’s Displacement Law: _peak  T  = constant = 2.898 x 10⁶ nm  K (or v­_max = constant * T)

Planck’s Law:

Wien: I() is a function of T so the peak occurs for a set value of T  peakT = constant. Stephan: change of variable

Ex 22.) Which of the following describes the effect of doubling the absolute temperature of a blackbody on its power output per square meter and on the wavelength where the radiation distribution is a maximum?

(A) The output power is increased by a factor of 16 and the maximum of the distribution shifts to twice its original wavelength.

(B) The output power is increased by a factor of 16 and the maximum of the distribution shifts to half its original wavelength.

C) The output power is increased by a factor of 8 and the maximum of the distribution shifts to twice its original wavelength.

(D) The output power is increased by a factor of 8 and the maximum of the distribution shifts to half its original wavelength.

(E) The output power is increased by a factor of 2 and the maximum of the distribution shifts to four its original wavelength.

Higher temperature suggests higher energy and hence shorter wavelength. Answers (B) and (D) are viable! The power increase factors of 2⁴ or 2³ for (B) and (D). The fourth power should be familiar as it is Stephan’s law, Power  T ⁴. Wien: _peakT = constant. Doubling the temperature moves the peak to one half of the wavelength. B

1 eV  1240 nm

1 eV  photon  = 1240 nm  2 eV  photon  = 620 nm; 4 eV  photon  = 310 nm

hc = 197 eV·nm (atomic physics) = 197 MeV·fm (nuclear physics)

color  (nm) v (1012 Hz) E(eV) IR > 700 < 428 < 1.77 red 625-700 405-480 1.77-2.0 orange 585-620 484-513  yellow 585-570 513-526  green 570-505 526-594  blue 500-440 600-682  violet 440-400 682-750  UV < 400 > 750 > 3.10 ranges are approximate and arbitrary no universal agreement on the ranges

Ex 23.) Let H denote a Hermitian operator and suppose that H |= a |, where | is an eigenvector of H. Which of the following is true of the eigenvalue a? (The symbols Re and Im denote the real and imaginary parts, respectively.)

(A) Re(a) = Im(a)

(B) Re(a) = - Im(a)

C) Re(a) = 0

(D) Im(a) = 0

(E) H = a

SP352 Knowledge points: The eigenvalues of a Hermitian operator are real. The eigenvectors associated with distinct eigenvalues are orthogonal. A Hermitian operator can be diagonalized by a unitary transformation. D

Ex 24.) The quantum numbers used to label the radial wave function solutions to the Shroedinger equation for hydrogen atom are the principal quantum number n and the angular momentum quantum number . If the principal quantum number is n = 2, which of the following gives the possible values for the angular momentum quantum number ?

(A) 1, 0 (B)  (C) 2, 1, 0 (D)  (E) ³/₂, ¹/₂

The states of the hydrogen atom are labeled by the quantum numbers n,  and m plus the intrinsic spin. The allowed values are n = 1, 2, 3, ….;  = 0, 1, … , (n -1); m_ = -, - +1, … , 0, 1, …, ).

That is: m_ assumes the 2+ 1 values spaced by 1 running from - to . Note: n and  are 

For n = 2,  can be 0 or 1 and m_ can be -1, 0, +1. A Each combination can be combined with spin up or down. The value  is the orbital angular momentum which assumes integer values. The total angular momentum is the orbital plus spin, and it assumes half-integer values for an electron in hydrogen.
Ex 25.) The three operators (L_x, L_y, L_z) for the components of the angular momentum commute with the Hamiltonian of a particular particle. Therefore, the angular momentum of the particle is

(A) equal to zero

(B) equal to the energy in magnitude

(C) always equal to L_z

(D) a unit vector

(E) a constant of the motion

The Hamiltonian is the energy and time-development operator for a quantum system. It results that

for time-independent operators. Quantities with time-independent operators that commute with the Hamiltonian are constants of the motion. E A quantum state can be labeled with good quantum numbers corresponding to it eigen-energy quantum number and quantum numbers each additional member of the largest set of mutually commuting operators that includes the Hamiltonian. For example, the members of the set (L_x, L_y, L_z) do not commute with one another. They do commute with L² = L_x² + L_y² + L_z². Considering that it is a given that (L_x, L_y, L_z) commute with the Hamiltonian, then L² will also commute with the Hamiltonian. In such a case, there would be good quantum numbers corresponding to H, L² and one of the components of angular momentum. L­_z has the simplest representation in spherical coordinates so it is usually chosen. For the hydrogen atom problem, the quantum numbers n,  and m are associated with eigenvalues for H, L² and L­_z.

Ex 26.) A particle of energy E is in an eigenstate of the square well potential shown above, with wave function (x). Which of the following is a correct expression for the expectation value of x² for this particle?

(A)

(B)

(C)

(D)

(E)
The correct procedure to compute an expectation value is to slap the operator for the quantity between *(x) and (x) and then to integrate of the full range of x for which x) is non-zero. This square well has a finite depth so the particle penetrates into the classically forbidden region |x| > a. In this case the integrals must run over the full range (-,). E
SPECIAL TOPICS:
(27) The Pauli Matrices:

_x = ; _y = ; _z =

Ex 27.) Given the Pauli spin matrices shown above and the operators defined by ₊ = _x + i _y and _- = _x - i _y, which of the following is NOT correct.

(A) ₊ (B) ₊ (C) _- (D) _z (E) _z

_- = C is false

Method: Realize that you have no idea what the problem is about, but also realize that it just requires some matrix multiplication. Find the explicit forms for ₊ and _-, and brute force compute each matrix product until you find the one that is NOT correct.
Background: ** Skip to the next problem if you are in a hurry.  (28) Counting Statistics

Suppose that a set of 2 x 2 matrices is sought such that any 2 x 2 matrix can be represented as a sum of the members of that set. Clearly the set must have at least four members as 2 x 2 matrices have four independent elements. One choice that could be made is:

This set of matrices is not very exciting. Spice it up; add the requirement that each member of the set be Hermitian (equal to the complex conjugate of its transpose). The simplest matrix that has imaginary elements that meets this requirement is:

An independent off-diagonal matrix needs to have elements that are equal rather than the negative of one another. Equal off-diagonal elements and Hermitian restricts the elements to be real.

Following the equal and negative scheme for the on diagonal matrices, the remaining members are:

and

This particular set of 2 x 2 matrices is the set of Pauli matrices. They have assigned labels:

1 = ; _x = ₁ =; _y = ₂ =; _z = ₃ =

The first matrix is the identity, and the final three are the Pauli matrices. The set of the four matrices is a basis for the collection of all 2 x 2 matrices.

An arbitrary 2 x 2 matrix can be represented as a linear combination of the four matrices, but not as linear combination of any set with fewer than four members. (A basis must be complete in the sense that all elements (matrices) of interest can be represented as linear combinations of the members of the set and economical in the sense that every member is necessary. If even one member of the set is removed then there will be at least one 2 x 2 matrix that cannot be represented as a linear combination of the remaining members.)

Why are the Pauli matrices introduced?

The Pauli matrices are introduced to act on two-row column matrices. Matrices of the forms:

, and . The column vectors and are a basis set for all the. The column vectors and are to be called (spin) UP () and (spin) DOWN ().
If you are not familiar with modern physics, skip this paragraph.
In quantum mechanics, the Pauli matrices add the flexibility that allows a wavefunction to represent the spin character of an electron. A two row column vector is appended to a function of position and time.

The column vector represents spin up while represents spin down. The spin part is to be normalized independently so:.

Special properties of the Pauli matrices:

₁_{1 =} ₂ _{2 =} ₃_{3 = 1}

The determinants are each of the three Pauli matrices is – 1. The determinant of the identity is, of course, equal to one.

|₁| = |₂| = |₃|= 1

Actions of the Pauli Set:

_x = ; _x =

The operator _x lowers the UP state to DOWN and raises the DOWN state to UP.

_y = ; _y =

The operator _y lowers the UP state to i times the DOWN and raises the DOWN state to –i times UP.

_z = ; _z =

The operator _z on the UP state to 1 times the UP and, on the DOWN state, returns – 1 times the DOWN state. The operator _z measures the up or down character of the state.
The combination of operators ₊ = _x + i _y annihilates UP and raises DOWN to 2 times UP.

₊ = ; ₊ =

The operator ₊ is called the raising operator.
Exercise: Find the action of  ­_- = _x + i _y on the states UP and DOWN. Propose a name for  ­_-.
** The Pauli matrices are multiplied by ½  when they are to be used in anger. They become the operators for the components of the electron’s spin angular momentum.

s_z =½  _z = ½ ; s² =(½ )² [_x² + _y² +_z²] = ³/₄ ²

The eigenvalues of s_z are  ½ , and the eigenvalue of s² is ³/₄ ² = s(s + 1)²

(28) Counting Statistics

We will assume that the quick and dirty answers are enough to fake our way through the problem. Memorize: When a counting experiment yields N counts, the uncertainty in that number is N ^{½ (or a fractional uncertainty of} N ^-½  large count numbers are relatively more precise.). The value N ^½ is the expected standard deviation. If the counting experiment is repeated many times, the expected results are assumed to follow a standard (Gaussian) distribution.

68% of the time the result is within   95% of the time the result is within  2  99.7% of the time the result is within  3  The average count number is, and we assume the standard deviation  is N ½. ** Know: 68% within  and 95% within 

https://en.wikipedia.org/wiki/Normal_distribution

Ex. 28) An experimenter measures the counting rate from a radioactive source as 10,150 counts in 100 minutes. Without changing any conditions, the experimenter counts for one minute. There is a probability of about 15% that the number of counts recorded will be less than

(A) 50 (B) 70 (C) 90 (D) 100 (E) 110

The first line established the average count rate as 101.5 per minute. We expect 101.5 counts in one minute on average. The standard deviation of the count number in one minute is assumed to be the square root of that value or 10.075. There is a 68% probability that the counts falls between 101.5 – 10.075 and 101.5 + 10.075 or a 32% chance that the number of counts is outside this range. The counts can be high or low, so there is a 16% chance that the number of counts in a minute is less than -  = 101.5 – 10.075 = 91.42.  about 15% likelihood to be less than 90. C
We expect 101.5 counts means that if the one minute count were repeated a large number of times that various integer results would be recorded each time. The results 101 and 102 would occur about equally, and larger and smaller values would occur at a lower frequency. The values around occurring about 0.6 times as often as those around . , e^-½  0.6.
(29) Graphs and Basic Lab

x(t) = 0.3190 t 2 + 0.0013

Ex.29) A motion sensor is used to measure the position x versus time t for a cart traveling down a ramp. A spreadsheet is then used to make a linear fit to the plot of x vs. t, as shown in the graph to the left. The equation for the best fit line appears on the graph. Which of the following gives the acceleration of the cart?

^m/_s2 ^m/_s2 C^m/_s2 D^m/_s2 E^m/_s2
This one is a gift! The acceleration is the second derivative of x(t) with respect to t or 0.6380. Adding dimensions, x(t) = 0.3190 ^m/_s 2 t ² + 0.0013 m yielding a = 0.6380 ^m/_s 2. D

Whenever possible, work from the definition. You are always correct to do so.

Alternative: Compare with the kinematical equation x(t) = ½ a t² + v_ot + x_o. ½ a = 0.3190 ^m/_s 2

(30) Lagrangian and Hamiltonian Mechanics:
We restrict our treatment to the simple cases for which the constraints are good, the potentials are time independent and for which there is no dissipation.

The lagrangian L is T – U, the kinetic energy minus the potential energy. The lagrangian is a natural function of the coordinates q_i, the coordinate velocities and time t. The natural variables are the ones that you must use!

The Lagrange equations of motion are:

The signs and details can be recovered by considering: .

(recovered Newton’s 2^nd Law)

It is crucial to note that the time derivative is a total derivative while the others are partial derivatives.

The momentum conjugate (associated with) a coordinate is found as: . Using the example , . Another familiar result is which appears for the planar orbit problem for which .
Conservation or constants of the motion:



The momentum p­_i is a constant of the motion if the lagrangian does not depend on q_i, the coordinate conjugate to that momentum. Given: , there is no dependence of the angular coordinate  so the conjugate momentum is a constant of the motion.

In Hamiltonian mechanics, the coordinates and the momenta are the natural variables. The hamiltonian is defined to be H = - L. Here p_i is the momentum conjugate to q_i as defined based on the lagrangian above. For the simple case, , we find .  H(x, p_x) = . We find:

The hamiltonian must be expressed as a function of its natural variables, the coordinates and momenta, so the second form is the one that must be used. For time-independent problems (the type usually encountered), the hamiltonian represents the total energy of the system.

The equations of motion are: and . Using the example: ,

and

Constants of the Motion: The relation shows that p_i is a constant of the motion of the hamiltonian is independent of q_i. (Recall that we are restricting our attention to hamiltonians that do not depend explicitly on time.)

Ex. 30) Two masses m1 and m2 on a horizontal straight frictionless track are connected by a spring of spring constant k, as shown in the figure to the left. The spring is initially at its equilibrium length. If x1 and x2

give the displacements of the masses from their equilibrium positions, the lagrangian L for the system is given by which of the following? The dot denotes differentiation with respect to time.

Note that the generalized coordinates chosen (see the figure) are the displacements of each mass from equilibrium. This choice is standard choice. The lagrangian is to be the kinetic energy minus the potential energy. The kinetic energy is the sum of the kinetic energies of the two particles, T (or K) = , and the potential energy is ½ times the spring constant times (x­₂ – x₁)², the stretch of the spring squared.  U = ½ k (x₂ – x₁)². Note that it helps to roll the words around in your mind as you attempt to decipher a problem.

Conclusion: B
Note that four of the candidates identify the kinetic energy as: , and three identify the potential as ½ k (x₂ – x₁)². Only two have the negative sign for L = T – U. Possibility (E) includes the awe-inspiring reduced mass; it has the wrong sign for the potential. In addition, (E) depends only on the relative position and relative velocity. That is: the center of mass motion has been suppressed. For example, if q = x₂ – x₁, then (E) is ½ + ½ k q². Forgetting that it should be ½ - ½ k q², there is the issue of the kinetic energy.

= + ½  where is the center of mass velocity.
Kinetic energy in terms of the center of mass and relative coordinates.
Let be the velocity of the center of mass of a system of particles, be the velocity of particle i

and be the velocity of particle i relative to the center of mass.

For a system with only two particles, the reduced mass can be used to get the special form:

where M = m₁ + m₂, and

Note that for rigid bodies the motion decomposes into the center of mass motion plus rotation about the CM.
(31) Nuclear Decays

Note that the inset image expands the energy scale.

Each nucleus has Z protons (aka atomic number) and N neutrons with a mass number of A =Z+N. A particular nucleus is denoted ; such as , however the Z and N are usually omitted because they are redundant and are known from A and the symbol used for X.

particle	charge	nucleon #	lepton #
nucleus ZXA	Z	A	0
photon 	0	0	0
alpha 	2	4	0
beta minus e-	-1	0	+1
beta plus e+	+1	0	-1
electron-neutrino ve	0	0	+1
electron-antineutrino	0	0	-1

An alpha particle is a tightly bound system of 2 protons and 2 neutrons (A=4). The very high total binding energy (~28 MeV; 7 MeV per nucleon) of the alpha and its very high first internal excited state make it favorable to be treated as a unit object. (The same could be said for the proton and neutron which in reality are quark composites.) The maximum binding energy per nucleon occurs for ⁵⁶Fe and is about 8.8 MeV. Elements to the left can in principle fuse to form ⁵⁶Fe and those to the right can fission.

Nuclei will spontaneously decay into another nucleus if the final system has lower rest energy than the initial. There are four decay modes; gamma decay, beta decay, alpha decay, and fission. The conserved quantities are total energy, linear and angular momentum, along with nucleon number and lepton number. The rest-mass energy difference between the initial and final systems is called the Q-value and appears as kinetic energy and internal excitation energy of the products.
In gamma decay, an excited nucleus decreases its level of excitation by emitting a gamma ray (photon)The process may be denoted: ^A(X*)  ^AX + ⁰ or just ^AX +  The number of neutrons and protons is unchanged. No leptons are involved so lepton number need not be considered.

(Leptons: electron, positrons and neutrinos  and . -- for nuclear)

In alpha decay, a nucleus emits an alpha particle (2 protons and 2 nucleons for a total of 4 nucleons). The final state nucleus, the daughter, has Z–2 protons, N-2 neutrons and A 4 nucleons. This can be denoted X  Y +  or just ^A-4Y + . No leptons are involved so lepton number need not be considered.

m_n =1.675 x 10^-27 kg = 939.57 ^MeV/­_c2;

m_p =1.672 x 10^-27 kg = 938.27 ^MeV/­_c2

m_e =9.11 x 10^-31 kg = 0.511 ^MeV/­_c2

There are three fundamental processes in beta decay. For convenience they can be considered to be (with Q-values given).

beta minus decay

beta plus decay

electron capture

Beta minus decay can occur for an isolated neutron, but because the Q-values for the remaining two processes are negative, the lower two reactions can only occur in regions where the proton is ‘interacting with an energetic background’; such as in a nucleus. Electron capture occurs when a ‘proton in the nucleus grabs an electron out of its s-state orbit’.

Note that the reactions above conserve charge, nucleon number, and lepton number.

For nuclei undergoing beta decay, the reactions would be written,

beta minus decay X  Y +e^ or just Y + e^. +

beta plus decay X  Y + e^_e

electron capture X   Y + _e

Note that the reactions above conserve charge, nucleon number, and lepton number. The existence of three-bodies in the final state means that the available energy or Q value of the reaction is not apportioned to the particles in a set ratio, but rather the particles of each type emerge with a distribution of energies.
In fission, the nucleus breaks into two large pieces (fission fragments) and a few neutrons with Q-values on the order of 200 MeV. A commonly used radioactive fission source is ²⁵²Cf. The nuclei ²³⁵U and ²³⁹Pu do not themselves spontaneously fission, but fission can be induced by striking them with a neutron. The conserved quantities are charge, total energy, linear and angular momentum, along with nucleon number and lepton number
Ex. 31) Thorium with atomic mass 228, decays by alpha emission to a daughter nucleus which also decays by alpha decay to radon. Which of the following is true of the decay product, radon? (The atomic number of thorium is 90.

	Atomic Mass	Atomic Number
(A)	220	82
(B)	220	86
(C)	224	82
(D)	224	88
(E)	228	91

The process is a sequence of two alpha decays:

X  Y +  Y + + 

Atomic Mass: A = 228  A_final – 4– 4

Atomic Number: Z = 90  Z_final = 90 – 2 – 2 = 86 

²²⁸Th  ²²⁴Ra +  ²²⁰Rn + + 

Thorium  Radium + alpha Radon + alpha + alpha

(32) Spontaneous Particle Decays:
As with other decays, the electric charge, energy, linear and angular momentum must be conserved. Because there is only one initial particle, the Q-value for the reaction must be positive. The excess energy or Q appears as kinetic energy and excitation energy of the products. Anti-particles have the same rest energy as their corresponding particles.
Muons and taus are heavy electrons each with its associated neutrino and anti-neutrino. Conserve electron, muon and tau lepton numbers separately! m_e = 0.511 ^MeV/_c2; m_ = 206 m_e; m_ = 3478 m_e
m_pc² = 938.27 MeV * m_nc² = 939. 57 MeV * m_nc² - m_pc² = 1.3 MeV > m_ec²

m_ec² = 0.511 MeV ^* m_c² = 106 MeV ^* m_c² = 1775 MeV m_c² = 0 *

m_c² = 140 MeV charged; 135 MeV neutral m_c² = 0 *

Be aware of the values marked * and that a pion is a little more massive than a muon.

Ex. 32) Which of the following decays is possible in a vacuum?

(A) ⁺  ^- + _

(^-  ⁺ + e^- + e⁺ 

(C)⁰  e^- + p 

(D)p  n + e⁺ + _e 

(E)n  p + e^- +

(A) Fails. Does not conserve charge. Q-value > 0 is OK.

(B) Fails for two reasons. Does not conserve charge. Does not conserve energy because Q<0, that is, the products have more mass than the initial particle.

(C) Fails. Conserves charge, however the products have much more massive than the pion. Electron lepton number is not conserved.

(D) Fails. Conserves charge and electron lepton number, however, the neutron is slightly more massive than the proton. The mass of the products is greater than the mass of the reactants  not a spontaneous decay.

(E) Possible. Conserves charge and the Q > 0. An isolated neutron is unstable and decays as shown with a lifetime of 886 seconds. Neutrons are stabilized in the nucleus by their binding energy which is greater than the Q of decay for an isolated neutron. E
Because the fundamental building blocks are the 6 quarks (u,d,s,c,b,t), 6 leptons (e,v_e,,v_,,v_) and their anti-particles, there are actually additional conservation rules. These are not as trivial to apply because one must be told the quark content of the composite particles.
Review the quark model section of your modern physics text if you have time.

http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/quark.html

Quarks and Leptons are the building blocks which build up matter, i.e., they are seen as the "elementary particles". In the present standard model, there are six "flavors" of quarks. They can successfully account for all known mesons and baryons (over 200). The most familiar baryons are the proton and neutron, which are each constructed from up and down quarks. Quarks are observed to occur only in combinations of two quarks (mesons), three quarks (baryons), and the recently discovered particles with five quarks (pentaquark).

Quark	Symbol	Spin	Charge	Baryon Number	S	C	B	T	Mass*
Up	U	1/2	+2/3	1/3	0	0	0	0	360 MeV
Down	D	1/2	-1/3	1/3	0	0	0	0	360 MeV
Charm	C	1/2	+2/3	1/3	0	+1	0	0	1500 MeV
Strange	S	1/2	-1/3	1/3	-1	0	0	0	540 MeV
Top	T	1/2	+2/3	1/3	0	0	0	+1	174 GeV
Bottom	B	1/2	-1/3	1/3	0	0	-1	0	5 GeV

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