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Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions. Other forms of nuclear matter are also studied.
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NUCLEAR PHYSICS
Asst.Prof., Department of Physics, Bhattadev University, Bajali
MSc. 2nd Semester Class notes
Everything we can see in the night time sky is made of nuclear matter. Nuclear physics describes how the Sun generates the energy we need for life on Earth, how all the atoms in your body were made in stars and what happens in stars when they die. Nuclear physics research tries to answer the fundamental questions: Where Do We Come From? What Are We? Where Are We Going? Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions. Other forms of nuclear matter are also studied. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons. The history of nuclear physics as a discipline starts with the discovery of radioactivity by Henri Becquerel in 1896 while investigating phosphorescence in uranium salts. In the years that followed, radioactivity was extensively investigated, notably by Marie and Pierre Curie as well as by Ernest Rutherford and his collaborators. By the turn of the century physicists had also discovered three types of radiation emanating from atoms, which they named alpha, beta, and gamma radiation. The 1903 Nobel Prize in Physics was awarded jointly to Becquerel for his discovery and to Marie and Pierre Curie for their subsequent research into radioactivity. Rutherford was awarded the Nobel Prize in Chemistry in 1908 for his ”investigations into the disintegration of the elements and the chemistry of radioactive substances”. The key experiment behind this announcement was performed in 1910 at the University of Manchester: Ernest Rutherford’s team performed a remarkable experiment in which Geiger and Marsden under Rutherford’s supervision fired alpha particles (helium nuclei) at a thin film of gold foil. Rutherford’s analysis of the data in 1911, led to the Rutherford model of the atom, in which the atom had a very small, very dense nucleus containing most of its mass, and consisting of heavy positively charged particles with embedded electrons in order to balance out the charge. That’s where the entire things start developing out.
As you sit on your chair, reading this article, with your laptop or desktop and your android phone nearby, you may be unaware of the many forces acting upon you. A force is defined as a push or pull that changes an object’s state of motion or causes the object to deform. Newton defined a force as anything that caused an object to accelerate according to F = ma, where F is force, m is mass and a is acceleration. The familiar force of gravity pulls you down into your seat, toward the Earth’s center. You feel it as your weight. Why don’t you fall through your seat? Well, another force, electromagnetism, holds the atoms of your seat together, preventing your atoms from intruding on those of your seat. The remaining two forces work at the atomic level, which we never feel, despite being made of atoms. The strong force holds the nucleus together. Lastly, the weak force is responsible for radioactive decay, specifically, beta decay where a neutron within the nucleus changes into a proton and an electron, which is ejected from the nucleus. Thus there are four fundamental forces present in nature. Let’s now become a bit more technical about these forces
Let’s now find how much 1 amu does in MeV corresponds to. This oftens comes in your exam with 2 marks. Here is how we do it. Well obviously using Einstein’s formula E = mc^2
E = mc^2 = 1. 6605 × 10 −^27 kg × (2. 9979 × 108 )^2 (m/sec)^2 = 15. 0639 × 10 −^11 J =
= 931. 5 M eV
As we know that 1. 6022 × 10 −^19 J = 1 eV. Now paste this in your memory chip. 1 amu = 931.5 MeV Let us take an example for the α-particle which is the 2 He^4 nucleus. This nucleus contains 2 protons and 2 neutrons. Now sum of the masses of the indivisual constituents can be calculated as follows
M′^ = 2 × 1 .007276 amu + 2 × 1 .008665 amu = 4.031882 amu
And the mass (M) of the α-particle is 4.001506 amu. That means we are in a position to calculate the mass defect of the nucleus which is ∆m = 4.031882 amu − 4 .001506 amu = 0.030376 amu. Hence the binding energy will be ∆ E = ∆m × 931 .5 MeV/amu = 0.030376 amu × 931 .5 MeV/amu = 28.29 MeV. Well this is the amount of energy gets released from one single He^4 nucleus. Now if I take 4 gms of He^4 it will contain Avogadro’s number of nuclei. Then the release of energy will be 28.29 MeV × 6. 023 × 1023 M ol−^1 which is almost, after doing a little bit of algebra,
Nuclear Stability is a concept that helps to identify the stability of a nuclear species. For example two isotope can have different abundance means their stability is different in different ways. But before we move on let us think about the following question. Why does every nucleus wants to get stability? Think of yourself when you are tired and ready for sleep. In this case you will most likely just stay put and not do anything as if you don’t have anything to spare. The major underlying reason is: ”Nature seeks the lowest energy state”. In the lowest energy state, things are most stable and less likely to change. One way to view this is that energy makes things happen. If a nucleus is at its lowest energy state, it has no energy to spare to make a change occur. The following information that talks about stability is all based on the nucleus tending towards the lowest energy state. Unstable nuclei will try and become stable by getting to a lower energy state. They will typically do this by emitting some form of radioactivity and change in the process. The main factors that determine nuclear stability are
pf =
Isotopic Mass − Mass Number Mass Number
Figure 1.1: Graph of Binding energy per unit nucleon vs mass number.
Important features of the graph: Few things we can interprete from the above graph which are indeed very important observation. Following are those
When the energy of the incident α-particle energy becomes too large, there is a deviation from the Rutherford scattering formula is observed. The reason for this is that the Rutherford scattering formula was derived assuming that the nucleus was a point particle. In reality it has a finite size with a radius R of order 10−^15 m. The nucleus therefore has a charge distribution, ρ(r). In terms of quantum mechanics we can write
ρ(r) = Ze |Ψ(r)|^2 ( where symbols have their usual meaning)
Nuclear ‘radius’ is the extent over which the electric charge distribution of the proton, and therefore its wavefunction, is not too small, although in principle the wave-function extends throughout all space. So in order to get a more detailed picture we use high energy electrons instead of α-particle to probe the charge distribution. We know the Rutherford’s scattering formula where the scattering of α- particles from nuclei can be modeled from the Coulomb force and treated as an orbit. The scattering process can be treated statistically in terms of the cross-section for interaction with a nucleus which is considered to be a point charge Ze. For a detector at a specific angle with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula:
In terms of scattering cross-section
dσ dΩ
Z e^2 4 π ǫ 0 KE
sin^4 θ 2
& In terms of no. N (θ) =
Ni n L Z^2 k^2 e^4 4 π^2 K E^2 sin^4 θ 2
where θ is the scattering angle, Ni is number of incident α-particle, n is the number of target atoms per unit volume, L is thickness of the target, Z is the atomic number of target atom, k is Coulomb’s constant, e charge of the electron and KE is the kinetic energy of the α-particle. For electrons which are moving faster ie relativistically with a velocity v close to c, there has to be a correction to be introduced and was first calculated by Mott and we have
dσ dΩ
M ott
dσ dΩ
Rutherf ord
v^2 c^2
sin^2
θ 2
The cross section for scattering from a point-like target is given by the Rutherford scattering formula. If the target has a finite spatial extent, the cross section can be divided into two factors, the cross section times the squared of a term called form factor, which takes care of the spatial extent and shape of the target. Thus the probability amplitude for a point-like scatterer gets modified by the form factor. So we have
dσ dΩ
exp
dσ dΩ
M ott
|F (p^2 )|^2
here p is the momentum transfered by the electron in the scattering and its magnitude is related to the scattering angle. Thus mathematically the form factor is defined as the ratio by which the scattering cross-section is reduced when the charge +Ze is spread out over a finite volume. In Coulomb scattering, the particular property of the spatial extent sampled is the charge distribution ρ(r) for the object for which there will be an potential V (r). Let us now find the expression of the form factor. In order to get to this we will use the first order Born approximation method (Recall your last semester quantum mechanics class). To first order (and up to a normalization) the Born Approximation the scattering amplitude can be written as
f (^) Born^1 st =
Ψf (k)
∣ (^) V (r)
∣Ψi(k) 〉
=
Ψ⋆f (k) V (r) Ψi(k) d^3 r =
e−ikf^ r^ V (r) eiki^ r^ d^3 r assuming plane waves eikr
V (r) ei(ki^ −kf^ )r^ d^3 r =
V (r) e
ipr ℏ d^3 r
This result is still quite general; in order to proceed we will need to assume a specific form for the potential, V (r). We
can describe an extended charge distribution by Ze ρ(r) with
ρ(r) d^3 r = 1 and since we are dealing with a coulomb
potential ie the potential experienced by an electron located at r′^ is given by
V (r) = −
Z e^2 4 π ǫ 0
ρ(r) |r − r′|
d^3 r′
Substituting this potential into the general expression for the first Born Approximation to the scattering amplitudes we get
f (^) Born^1 st = −
Z e^2 4 π ǫ 0
ρ(r) |r − r′|
d^3 r′
e
ipr ℏ d^3 r
Now making the substitution r − r′^ = R and d^3 R = d^3 r
f (^) Born^1 st = −
Z e^2 4 π ǫ 0
ρ(r′) |R|
d^3 r′
e
ip(R+r′^ ) ℏ (^) d^3 R
Z e^2 4 π ǫ 0
e
ipR ℏ
|R|
d^3 R
ρ(r′) e
ipr′ ℏ (^) d^3 r′
Qualitatively, this can be interpreted as that the part of the wavefront that passes through the nucleus at a distance r from the centre and is scattered through an angle θ travels a further distance than the part of the wave that passes through the centre, by an amount proportional to r′^ and therefore suffers a phase change. This phase change also
depends on the scattering angle θ and is equal to pr′ ℏ
. This means that different parts of the wavefront suffer a different
phase change (just as in optical diffraction) these different amplitudes are summed to get the total amplitude at some scattering angle θ and this gives rise to the diffraction pattern. The contribution to the amplitude from the part of the wavefront which passes at a distance r from the centre of the nucleus is proportional to the charge density, ρ(r′) at r′. The total scattering amplitude is therefore the sum of the amplitudes from all these different parts, which is what the last integral means. Hence the bracked factor in the last equation is known as the ‘Form Factor’ F (p^2 ) and is defined by this integral over the volume of the target which is nothing but just the Fourier transform of the charge distribution of course in 3D. Since the Coulomb’s potential is spherically symetric therefore the form factor can be further simplified into
F (p^2 ) = 4 π ℏ Ze p
0
r′^ ρ(r′) sin
pr′ ℏ
dr′^ with d^3 r′^ = r′^2 dr′^ sinθdθ dφ
So an inverse Fourier transform of the form factor is then going to give us the charge distribution. Note here that the
form factor will be zero for a case when sin
pr′ ℏ
= 0 ie
pr′ ℏ = π. From here we can make a rough estimate of the
distance where the charge distribution changes from the order of its value at the centre to zero giving an approximate nuclear radius of about 3 fm. To be this value the charge distribution ρ(r) should take a form given by
ρ(r) =
ρ 0 1 + exp
( (^) r − R δ
where ρ 0 is the charge density at the center, R as the nuclear ‘radius’ and δ as the ‘surface depth’ which measures the range in r over which the charge distribution changes from the order of its value at the centre to much smaller than this value. This is known as the Fermi distribution. Sometimes also known as the Saxon-Woods model for charge distribution.
In 1922, at the University of Frankfurt (Germany), Otto Stern and Walther Gerlach, did fundamental experiments in which beams of silver atoms were sent through inhomogeneous magnetic fields to observe their deflection. These experiments demonstrated that these atoms have quantized magnetic moments that can take two values. Although consistent with the idea that the electron had spin, this suggestion took a few more years to develop.
operator on that space. If one builds a device to measure such an observable, and if one uses that device to make a measurement of that observable on the system, then the machine will output an eigenvalue λ of that observable. Moreover, if the system is in a state
, then the probability that the result of measuring that quantity will be the
eigenvalue of the observable is
λ
Suppose, now, that the system we are considering consists of the spin of a particle. The Hilbert space that models the spin state of a system with spin s is a 2s + 1 dimensional Hilbert space. Elements of this vector space are often called “spinors”. The cartesian components of the spin (which is an observable) of the system are three self-adjoint operators conventionally called Sx, Sy and Sz , as spin matrices whose eigenvalues are the possible values one might get if one measures one of these components of the system’s spin. More explicitly, since protons and neutrons are
spin-
system, and one chooses to represent states and observables in a basis consisting of the normalized eigenvectors
of the z component of spin, then one would find the following matrix representations in that basis
Sx =
σx Sy =
0 −i i 0
σy Sz =
σz
where σx, σy and σz are called as the Pauli spin matrices or Pauli spinors. These spinors have the follwoing properties Tr [σi] = 0
∣σi
∣ (^) = − 1 and σx σy σz = i
Since you know that sum of the eigenvalues of a matrix is equal to the trace of a matrix here a trace-less
matrix would indicate that the sum of the eigenvalues of the matrix must be zero. Thus you get spins are ±
. Also
a trace-less matrix would indicate a commutator. Since spin is a type of angular momentum, it is reasonable to suppose that it possesses similar properties to orbital angular momentum. Thus we can arrive at the same conclusion that [Sx Sy ] = i ℏ Sz [Sy Sz ] = i ℏ Sx [Sz Sx] = i ℏ Sy
In the similar note we can also have the matrix algebra of the Pauli spinors. Let’s just a look at it.
σx σy =
0 −i i 0
= i
= i σz
If you reverse the order ie σy σx will become −i σz. This in turn will lead to general conclusion that
σx σy = i σz = − σy σx σy σz = i σx = − σz σy σz σx = i σy = − σx σz
which have retained the its cyclic behaviour. So immediately we can have
σx σy − σy σx =
0 −i i 0
0 −i i 0
= 2 i
= 2 i σz
This result can be generalised as [σi σj ] = 2 i σk which have also retained its cyclic behaviour. Here we dfine a new object called commutator. Mathematically speaking the commutator of two operators is defined as the difference between the products of the two operators taken in alternate orders. Thus we have found that the Pauli spinors which do not commute since the value of the commutator is not zero. Had it been zero then AB − BA = 0 which will mean AB = BA. Thus for commuting operators, the order of operation does not matter, which then further will lead us to the fact that these two operators will have then simultaneous sets of eigenstates. Thus we say that we can know the eigenvalues of these two observables simultaneously. It is common to extend this language and say that these two observables can be measured simultaneously, though, we do not really measure them simultaneously. What we mean is that we can measure one observable without erasing our knowledge of the previous results of the other observable. Observables A and B are said to be compatible. Thus this is straight cut violation of Heisenberg’s uncertainty principle. But for non-commuting operators these all will not happen. Thus when it comes to spin the conclusion to draw from this is that while we can know one spin component absolutely ( ie ∆ Sz = 0 ), we can never know all three, nor even two, simultaneously. This lack of ability to measure all spin components simultaneously implies that the spin does not really point in a given direction, as a classical spin or angular momentum does. So when we say that we have measured “spin up,” we really mean only that the spin component along that axis is up, as opposed to down, and not that the complete spin angular momentum vector points up along that axis. Also it can be easily seen that
σx σy + σy σx = σy σz + σz σy = σz σx + σx σz = 0
This is known as “anti-commuatation”, i.e., not only do the spin operators not commute amongst themselves, but the anticommute! They are strange beasts.
Another indication that the spin does not point along the axis along which you measure the spin component is obtained by considering a new operator that represents the magnitude of the spin vector but has no information about the direction. It is common to use the square of the spin vector for this task. This new operator is
S^2 = S x^2 + S^2 y + S z^2
0 −i i 0
Thus the S^2 operator is proportional to the identity operator, which means it must commute with all the other operators Sx, Sy and Sz ie [S^2 , Si] = 0. It also means that all states are eigenstates of the S^2 operator.
Parity is a rather subtle concept and has no classical analogue. It is concerned with the behaviour of wave function under space inversion. Two kinds of parity actually correspond to two different kinds of quantum wave function for a particle. Nuclear states have a well defined parity, defined by the behaviour of the wavefunction for all the nucleons under reversal of their coordinates with the centre of the nucleus at the origin. This is equivalent to studying the mirror image of the original system. It was originally assumed that parity must be conserved in all particle interactions, but it was demonstrated that parity does not have to be conserved in β- decay. Parity of nuclear ground states can usually be determined from the shell model which be dealt in coming chapters.
Ψ (−x, −y, −z) = λ Ψ (x, y, z)
The parity operator Π is defined as Π
∣x〉^ =
∣ (^) − x〉^ which is a Hermitian such that Π†^ = Π. So the eigen value
operation is given by from definition
Π
∣Ψ〉^ = λ
∣Ψ〉^ = λ^2
= λ^2
= λ^2
λ^2 = 1 = λ ± 1
Thus if x represents space then we have
x
− x
= Ψ(−x) = λΨ(x). The parity is eventually
Ψ(x) =
even λ = 1 odd λ = − 1
Let me just draw an analogy to introduce the concept of “Isospin”. We already know that electrons have two
spin values with respect to the z-direction. ie Sz = ±
which then can be distinguished by the application of an
external non-uniform magnetic field in the z-direction. But in the absence of this external field these two cannot be distinguished and we are used to thinking of these as two states of the same particle. So we need to invoke the principle of superposition to describe the state of the electronic spin. Similarly, if we could ‘switch off’ electromagnetic interactions we would not be able to distinguish between a proton and a neutron. Also as far as the strong interactions are concerned it is also charge independent. So we just can’t distinguish protons and neutrons as a charged and neutral particle in nuclear physics. Thus then, these are just two states of the same particle (a nucleon). Then how will we distinguish between them? The answer is by isospin. What we therefore think of an imagined space in which the nucleon has this property which is mathematically analogous to spin but has nothing to do with angular momentum. The proton and neutron are now considered to be a nucleon with different values of the third component of this isospin I 3 or sometimes Iz. This isospin is associated with a conservation law which requires strong interaction decays to conserve isospin. This term was derived from isotopic spin, but physicists prefer the term isobaric spin, which is more precise in meaning.
Since this third component can take two possible values, we assign I 3 =
for the proton and I 3 = −
for the
In the last chapter we spoke about stability of nuclei. But not all the nuclei are stable. What does that lead to? This chapter will focus on what the unstable nuclei will do? Well it’s just simple. They will decay. Because the nucleus experiences the intense conflict between the two strongest forces in nature, it should not be surprising that there are many nuclear isotopes which are unstable and emit some kind of radiation. Radioactive decay (also known as nuclear decay or radioactivity) is the process by which an unstable atomic nucleus loses energy by emitting radiation, such as an α particle, β particle or γ particle. To put it in another way the atomic nuclei that dont have enough binding energy to hold the nucleus together due to an excess of either protons or neutrons are going to disintegrate. Let’s have some history. Radioactivity was discovered in 1896 by the French scientist Henri Becquerel, while working with phosphorescent materials. These materials glow in the dark after exposure to light, and he suspected that the glow produced in cathode ray tubes by X-rays might be associated with phosphorescence. He wrapped a photographic plate in black paper and placed various phosphorescent salts on it. All results were negative until he used uranium salts. The uranium salts caused a blackening of the plate in spite of the plate being wrapped in black paper. It soon became clear that the blackening was also produced by non-phosphorescent salts of uranium and metallic uranium. These radiations were given the name ”Becquerel Rays”.
The modes and characteristic energies that comprise the decay scheme for each radioisotope are specific. If instru- mentation is sufficiently sensitive, it is possible to identify which isotopes are present in a sample. But that will cost lot of your money. Radioactive decay will change one nucleus to another if the product nucleus has a greater nuclear binding energy than the initial decaying nucleus. The difference in binding energy (comparing the before and after states) determines which decays are energetically possible and which are not. But let me put all the information about radioactivity in a straight forward form.
The harsh reality is that radioactivity has not been invented by man; it has been there, existing in the universe since time immemorial. The of nuclei which takes place in nature, is called natural radioactivity. However there are elements beyond uranium which have been artificially made. They are called the transuranium elements which can be made to
β-particles are either electrons or positrons that are emitted through a certain class of nuclear decay associated with the weak force which is characterized by relatively lengthy decay times. The name β, followed naturally as the next letter in the Greek alphabet after α, α-particles having already been discovered and named by Rutherford. But as we know that the radioactivity is entirely a nuclear phenomenon then where does this e−^ come from? But can you remember that the neutron has a larger mass than the proton and is thus unstable with respect to the combination of a proton and an electron. So consider the following
1 n
1 p
0 e − 1
1 p
1 n
0 e
Thus inside the nucleus if these things happens it will result in a production of an e−. Aha..! We now know that there can be an e−^ production. But then again why does that produced e−^ comes out of the nucleus. I mean why it can’t stay inside the nucleus? To answer this we need to do a little bit of algebra using some celebrated principles of physics. Next is how we can show that. We know the Heisenberg’s uncertainty principle as ∆x ∆p ≥ 2 h π. Take ∆x as positional uncertainty which is equal the to typical nuclear dimension means the e−^ can be anywhere inside the nucleus. Thus ∆x = 10−^15 m. Mass of the e−=9. 1 × 10 −^31 kg. Now do a calculation.
∆x ∆p =
h 2 π ∆x m∆v =
h 2 π ∆v =
h 2 π m ∆x =
= 1. 2 × 1011 m sec−^1
That’s velocity at which the e−^ has to stay inside the nucleus which is straightway violating the Special Theory of Relativity according to which nothing can have a velocity greater the velocity of light. Thus Heisenberg’s uncertainty principle along with STR will speak about why can’t an e−^ reside inside nucleus. Hence the e−^ has to come out of course.
Proton decay, neutron decay, and electron capture are three ways in which protons can be changed into neutrons or vice-versa; in each decay there is a change in the atomic number, so that the parent and daughter nuclei are different. In all three processes, the number A of nucleons remains the same, while both proton number, Z, and neutron number, N, increase or decrease by 1. So far so good! Now let’s get somewhat detailed into that.
Z XA^ →^ Z+1Y^ A^ +^ e−^ + ¯νe
where A and Z are the mass number and atomic number of the decaying nucleus, and X and Y are the initial and final elements, respectively. Inside the nucleus following is what that has happened.
0 n
1 p (^1) + e− (^) + ¯ν e
In β- decay, the mass difference between the parent and daughter particles is converted to the kinetic energy of the daughter particles. This kinetic energy is of course coming from masses of atoms involved in process. Though the atomic mass is almost comparable with the nuclei but still there is minute difference since in case atom the electrons have to also taken into account and they also contribute to the mass. So we must concentrate the nuclear mass rather than the atomic mass. Since neutrinos are massless therefore neglecting it in the equation
A (^) + e−
the disintegration energy Q can be written down as
Q = [Nuclear mass (X)] − [Nuclear mass (Y ) + me− ] × c^2 = [Atomic mass (X) − Zme− ] − [Atomic mass (Y ) − (Z + 1)me− + me− ] × c^2 = [MX − Zme− − MY + Zme− + me− − me− ] × c^2 = [MX − MY ] in energy units
Thus for Q> 0 you must have MX > MY. Or to put it in a sentence “for β−-decay to occur the mass of parent atom must be greater than that of the daughter atom.”
Z XA^ →^ Z− 1 Y^ A^ +^ e+^ +^ νe
where A and Z are the mass number and atomic number of the decaying nucleus, and X and Y are the initial and final elements, respectively. Inside the nucleus following is what that has happened.
1 p
0 n (^1) + e− (^) + ν e
Similar treatment I am going to use. We will see the disintegartion energy pertaining to this decay. And will find out the condition. Hence in the equation
Z XA^ →^ Z− 1 Y^ A^ +^ e+
the Q value of the reaction is
Q = [Nuclear mass (X)] − [Nuclear mass (Y ) + me− ] × c^2 = [Atomic mass (X) − Zme− ] − [Atomic mass (Y ) − (Z − 1)me− + me− ] × c^2 = [MX − Zme− − MY + Zme− − me− − me− ] × c^2 = [MX − MY − 2 me− ] in energy units
Thus for Q> 0 you must have MX > MY + 2me−. Or to put it in a sentence “for β−-decay to occur the mass of parent atom must be greater than that of the daughter atom by at least twice the electronic mass.”
A (^) + e− (^) → Z− 1 Y^ A (^) + ν e
where A and Z are the mass number and atomic number of the decaying nucleus, and X and Y are the initial and final elements, respectively.
Here also the process is going to be same. But one thing is different in this case. See the electron was orbiting before it was getting captured by the nucleus. So it was as if pulled working against the binding energy of the electron in the orbit. So that energy has to be taken into account. The Q value of the reaction is
Q = [Nuclear mass (X) + me− ] − [Nuclear mass (Y )] × c^2 − Be = [Atomic mass (X) − Zme− + me− ] − [Atomic mass (Y ) − (Z − 1)me− ] × c^2 − Be = [MX − Zme− + me− − MY + Zme− − me− ] × c^2 − Be = [MX − MY ] × c^2 − Be
Thus for Q> 0 you must have MX > MY + Be. Or to put it in a sentence “for K-capture to occur the mass of parent atom must be greater than that of the daughter atom by at least the binding energy of the electron.”