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(k_M / ɛ) values for liquids at 20 °C (589.29nm)

Benzene	η = 1.501	kM / ε = 1.99 × 10^8
Carbon tetrachloride	η = 1.461	kM / ε = 2.05 × 10^8
Carbon disulfide	η = 1.628	kM / ε = 1.84 × 10^8
Ethanol (ethyl alcohol)	η = 1.361	kM / ε = 2.204 × 10^8
10% Glucose solution in water	η = 1.3477	kM / ε = 2.22 × 10^8
20% Glucose solution in water	η = 1.3635	kM / ε = 2.200 × 10^8
60% Glucose solution in water	η = 1.4394	kM / ε = 2.08 × 10^8
sucrose	η = 1.3344	kM / ε = 2.24× 10^8

Amount of radiant intensity absorbed,

I' = (I − I")

Since I" = I exp (– 2.303k_M η ℓ C/ c). Consequently we may write without further hesitation that

I' = I (1 − exp (– 2.303k_M η ℓ C/ c))

The fluorescence intensity (F) is proportional to the amount of radiant intensity absorbed:

F = I' Q = I φ (1 − exp (– 2.303k_M η ℓ C/ c))

where φ = fluorescence quantum yield. The fluorescence quantum yield (φ) gives the efficiency of the fluorescence process. It is defined as the ratio of the number of photons emitted to the number of photons absorbed. When (2.303k _M η ℓ C/c) < 0.05, which can be achieved at low concentrations of analyte, the fluorescence intensity can be expressed as:

F = (2.303I φ k_M η ℓ /c) C

At low concentrations, less than 10⁻⁵ M, fluorescence intensity is linear and proportional to concentration of analyte with slope = 2.303 I φ k_M η ℓ /c.

For substances other than solutions the absorbance is given by:

Absorbance = k" η ℓ /c

When discussing the mass extinction rate coefficient, this equation is rewritten:

Absorbance = (k_µ η ℓ /c) ρ

where ρ = density of absorbing chemical species and k_µ = mass extinction rate coefficient. The mass extinction rate coefficient is a measurement of how fast a chemical species absorbs light at a given wavelength, per unit mass. The molar extinction rate coefficient is closely related to the mass extinction rate coefficient by the equation:

Molar extinction rate coefficient = mass extinction rate coefficient × molar mass

k _M = k_µ × molar mass

At low densities, less than 10⁻³ g /cm³, absorbance is linear and proportional to density of absorbing chemical species with slope = k_µ η ℓ/ c.

“An experiment is a question which science poses to Nature, and a measurement is the recording of Nature’s answer.”

--Max Planck

Understanding the natural growth of tumors is of value in the study of tumor progression, along with that it will be supportive for a better assessment of therapeutic response. Patterns of tumor growth can be shaped by a variety of factors, and so cancer biologists have developed different mathematical expressions, or models, to describe tumor growth rate. One of the most basic models of tumor growth rate is the exponential model. The exponential model was introduced by Skipper et al. (1964) and has proven to be well suited to describe the early stages of tumor growth. In clinical studies, where the natural growth of tumor can be preceded for a restricted period, the exponential model is used to describe the growth of tumors. An untreated tumor growth is usually well approximated by an exponential growth model. Some studies have shown that tumor growth rate may descend with time, which results in non-exponential growth model of tumors. A number of non-exponential growth models are available in the literature, among which the Gompertz model is widely used. The Gompertz Model was introduced by Gompertz (1825) and has proven to be well suited to describe the growth of an unperturbed tumor. The current view of tumor kinetics is based on the general assumption that tumor cells grow exponentially. Such kinetics agrees with the early stages of tumor growth. The time it takes for the tumor cells to reach twice its number density, doubling time DT, is of value for quantification of tumor kinetics, along with that it will be supportive for optimization of screening programs, prognostication, optimal scheduling of treatment strategies, and assessment of tumor aggressiveness.

If the tumor volume = V₀ at time t = 0 and

= V at any time t, then according to exponential model

V = V₀ e ^{α t}, where α is an exponential growth constant characterizing the growth rate of tumor volume. This model implies that the tumor volume can increase indefinitely and the growth rate of tumor is proportional to its volume:

dV /dt = α V

Tumor volume doubling time:

DT₁ = 0.693/α

According to equation above, the variation of DT₁ with α is:

dDT₁ / dα = −0.693/α²

If the number of tumor cells = N₀ at time t = 0 and

= N at any time t, then according to exponential model:

N = N₀ e ^{β t}, where β is an exponential growth constant characterizing the growth rate of tumor cells.

This model implies that the tumor cells can increase indefinitely and the growth rate of tumor cells is proportional to its number:

dN /dt = β N

The time it takes for tumor cells to double in number, doubling time DT₂, is represented by the following equation:

DT₂ = 0.693/β

According to equation above, the variation of DT₂ with β is:

dDT₂ / dβ = −0.693/β²

Further, the ratio dN / dV can be given as per the following expression:

(dlnV / dlnN) = (α/ β)

or

dlnV = (DT₂ / DT₁) dlnN

On integration within the limits of V₀ to V for tumor volume and N₀ to N for number of tumor cells, we get

ln (V / V₀) = (DT₂ / DT₁) ln (N / N₀)

A plot of ln (V / V₀) versus ln (N / N₀) is expected to be a straight line passing through origin with slope = (DT₂ / DT₁). n is the number density of tumor cells, defined as:

n = N/V which on rearranging:

n × V = N

Differentiating with respect to N gives:

(dn/dN) V + (dlnV/dlnN) = 1

Since (dlnV/dlnN) = (DT₂ / DT₁). We have then

(dn/dN) V + (DT₂ / DT₁) = 1

or

(dn/dN) = (DT₁− DT₂) / V DT₁

or

(dn/dt) / (dN/dt) = (DT₁− DT₂) / V DT₁

But dN /dt = 0.693N/ DT₂ and hence, we get

(dn/dt) = 0.693 (DT₁− DT₂) n / DT₂ DT₁

or

(dlnn) = {0.693 (DT₁− DT₂) / DT₂ DT₁} dt

On integration within the limits of n₀ to n for number density of tumor cells and 0 to t for the time, we get

ln (n/ n₀) = {0.693 (DT₁− DT₂) / DT₂ DT₁} t

The time it takes for the tumor cells to reach twice its number density, doubling time DT, is of value in the study of tumor progression, along with that it will be supportive for optimal scheduling of treatment strategies. Now, n = 2n₀ at t = DT. Therefore, equation above can be rewritten as follows:

DT = (DT₁− DT₂) / DT₂ DT₁

Assuming that tumor growth is exponential, the following equation is justifiable:

DT = DT₁ DT₂/ (DT₁− DT₂), where DT is the time it takes for the tumor cells to reach twice its number density, DT₁ is the time it takes for a tumor to double in volume, and DT₂ is the time it takes for tumor cells to double in number. This equation predicts the following limiting possibilities.

If DT₁ = DT₂, then

DT = ∞ which means: that it takes an infinitely long time for the tumor cells to reach twice its number density.

And according to the equation: dlnV = (DT₂ / DT₁) dlnN if DT₁ = DT₂ then dlnV = dlnN which means: V is proportional to N i.e., number density of tumor cells remains constant.

If DT₁ = DT₂, then number density of tumor cells remains constant and it takes an infinitely long time for the tumor cells to reach twice its number density.

If DT₁ >> DT₂, then

DT= DT₂

i.e., the time it takes for the tumor cells to reach twice its number density equals the time it takes for tumor cells to double in number.

If DT₂ >>DT₁, then

DT = − DT₁

i.e., because of negative sign the actual value of DT will be = 1 /DT₁.

The physicist Leo Szilard once announced to his friend Hans Bethe that he was thinking of keeping a diary: "I don't intend to publish. I am merely going to record the facts for the information of God."

"Don't you think God knows the facts?" Bethe asked.

"Yes," said Szilard.

"He knows the facts, but He does not know this version of the facts."

−Hans Christian von Baeyer, Taming the Atom

A Relativistic Bohr Model

“It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15- inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive center, carrying a charge.”

- Ernest Rutherford

According to the law that nothing may travel faster than the speed of light – i.e., according to the Albert Einstein’s law of variation of mass with velocity (the most famous formula in the world. In the minds of hundreds of millions of people it is firmly associated with the menace of atomic weapons. Millions perceive it as a symbol of relativity theory):

m = m₀ / (1− v²/c²) ^½

or

m²c² – m²v² = m₀²c²

That the mass m in motion at speed v is the mass m₀ at rest divided by the factor (1− v²/c²) ^½ implies: the mass of a particle is not constant; it varies with changes in its velocity.

Differentiating the above equation, we get:

mv dv + v²dm = c²dm

or

dm (c² – v²) = mv dv

In relativistic mechanics (the arguably most famous cult of modern physics, which has a highly interesting history which dates back mainly to Albert Einstein and may be a little earlier to H. Poincaré), we define the energy which a particle possess due to its motion i.e., kinetic energy to be = dmc² = dp × v. Therefore:

dp (c² – v²) = mc² dv

or

(dp/dt) = mc² / (c² – v²) (dv/dt)

Since: (dp/dt) = F (force) and (dv/dt) = a (acceleration), therefore:

F = mac² / (c² – v²)

(Note: For non-relativistic case (v << c), the above equation reduces to F = m₀a)

Because

m = m₀ / (1− v²/c²) ^½ or c² / (c² – v²) = m²/m₀². Therefore:

F = m³a / m₀²

Bohr Model:

In 1911, fresh from completion of his PhD, the young Danish physicist Niels Bohr left Denmark on a foreign scholarship headed for the Cavendish Laboratory in Cambridge to work under J. J. Thomson on the structure of atomic systems. At the time, Bohr began to put forth the idea that since light could no long be treated as continuously propagating waves, but instead as discrete energy packets (as articulated by Planck and Einstein), why should the classical Newtonian mechanics on which Thomson's model was based hold true? It seemed to Bohr that the atomic model should be modified in a similar way. If electromagnetic energy is quantized, i.e. restricted to take on only integer values of hυ, where υ is the frequency of light, then it seemed reasonable that the mechanical energy associated with the energy of atomic electrons is also quantized. However, Bohr's still somewhat vague ideas were not well received by Thomson, and Bohr decided to move from Cambridge after his first year to a place where his concepts about quantization of electronic motion in atoms would meet less opposition. He chose the University of Manchester, where the chair of physics was held by Ernest Rutherford. While in Manchester, Bohr learned about the nuclear model of the atom proposed by Rutherford. To overcome the difficulty associated with the classical collapse of the electron into the nucleus, Bohr proposed that the orbiting electron could only exist in certain special states of motion - called stationary states, in which no electromagnetic radiation was emitted. In these states, the angular momentum of the electron L takes on integer values of Planck's constant divided by 2π, denoted by ħ = h/2π (pronounced h-bar). In these stationary states, the electron angular momentum can take on values ħ, 2ħ, 3ħ... but never non-integer values. This is known as quantization of angular momentum, and was one of Bohr's key hypotheses. For circular orbits, the position vector of the electron r is always perpendicular to its linear momentum p. The angular momentum L = p × r has magnitude L = pr = mvr (where m = m₀ / (1− v²/c²) ^½, m₀ = rest mass of the electron) in this case. Thus Bohr's postulate of quantized angular momentum is equivalent to

mvr = nħ

where n is a positive integer called principal quantum number. It tells us what energy level the electron occupies. For an electron to orbit the nucleus in the circular orbit of radius r, it should obey the condition:

Ze²/4πε₀r² = m³a / m₀²

Since a = v²/r. Therefore:

Ze²/4πε₀r = m³v² / m₀²

or

Ze²/4πε₀ = m²v (mvr) / m₀²
Since: mvr = nħ

Ze²/4πε₀ = nħ m²v / m₀²

or

v = (Ze²/ 4πε₀nħ) × (m₀² / m²)

Note: for nonrelativistic model (i.e., v << c) the above expression reduces to:

v = (Ze²/ 4πε₀nħ)

Expression for radius of the orbit:

Substituting v = (Ze²/ 4πε₀nħ) × (m₀² / m²) in the equation mvr = nħ, we get:

r = (4πε₀n² ħ² / Ze²) × (m /m₀²)

Note: for nonrelativistic model (i.e., v << c) the above expression reduces to:

r = (4πε₀n² ħ² / m₀ Ze²)

Expression for potential energy of the electron:

For an electron revolving in n^th orbit of radius r

Potential energy is given by:

E_P = (potential at a distance r from the nucleus) × (-e)

E_P = (Ze/4πε₀r) × (- e) where Z is the atomic number and –e the charge on the electron.

i.e., E_P = – Ze² /4πε₀r

Substituting r = (4πε₀n² ħ² / Ze²) × (m /m₀²), we get:

E_P = – m₀²Z² e⁴ /16π²ε₀² n² ħ² m

Note: for nonrelativistic model (i.e., v << c) the above expression reduces to:

E_P = –m₀Z² e⁴ /16π²ε₀² n² ħ²

This energy represents the binding energy of the electron. Binding energy of an electron is the minimum energy required to knock out an electron from the atom. It is also denoted by E_B i.e., E_B = –m₀Z² e⁴ /16π²ε₀² n² ħ²

If a photon energy hυ is supplied to remove the electron from the n^th orbit, then this energy should be = E_B i.e., the condition hυ = – m₀²Z² e⁴ /16π²ε₀² n² ħ² m should be satisfied. If hυ is < than – m₀²Z² e⁴ /16π²ε₀²n²ħ²m, then it is impossible to remove an electron from the atom.

m²c² – m²v² = m₀²c²

Since: v = (Ze²/ 4πε₀nħ) × (m₀² / m²)

Therefore, the above equation becomes:

m²c² – (Z²e⁴ m₀⁴/ 16π²ε₀²n²ħ² m²) = m₀²c²

or

m²c² = m₀² {c² + Z²e⁴ m₀²/ 16π²ε₀²n²ħ² m²}

or

m²c² / m₀² = {c² + Z²e⁴ m₀²/ 16π²ε₀²n²ħ² m²}

For non-relativistic case (i.e., v << c)

m = m₀

c² = {c² + Z²e⁴ / 16π²ε₀²n²ħ²}

From this it follows that

Z²e⁴ / 16π²ε₀²n²ħ² = 0 (which is an illogical and meaningless result)
It is not good to introduce the concept of the mass M = m / (1– v²/c²) ^½ of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.
—Albert Einstein in letter to Lincoln Barnett, 19 June 1948