Thus, from Equation (5)

c ′ = λ 1 ν ′ 1 = π R s 2 meter / kshana (19)

where ν' cycles/kshana is the frequency of gamma radiation, and R_s meters is the radius of the spinning electron or positron. T ′ 1 = 1 / ν ′ kshanas is the time period of gamma radiation , which is also equal to the period of the spinning electron. The attributes of gamma rays are assumed to be present in electrons or positrons based on the thoughts of Maharishi Kanada [6].

c ′ = λ 1 T ′ 1 = π R s 2 meter / kshana (20)

However, from the definition of kshana, the period of spinning electron is T' = 4 kshanas. Therefore

λ 1 4 = π R s 2 meter / kshana (21)

Thus,

R s = λ 1 2 π meter (22)

Substituting the value of λ₁, we obtain the radius of the spinning electron as R_s = 3.8615926758 × 10⁻¹³ meters. The radius R_s can also be found by the law of conservation of energy in the pair production, which is hν₁ = m₀c² = 0.51 MeV [13]. Thus,

h ν 1 = h c λ 1 = 0.51   MeV (23)

h c 2 π R s = 0.51   MeV (24)

From Equation (22) substituting the value of λ₁ and the values of other constants in Equation (24), we have

R s = Λ ′ C = 3.86915646858 × 10 − 13     meter (25)

The spinning electron radius (R_s) calculated in Equations (22) and (25) are exactly equal to the reduced Compton wavelength Λ ′ C = 3.8615926796 (12) × 10⁻¹³ m [10].

Substituting the value of electron radius from Equation (25) in Equation (9), we have

n = 1.90853806367 × 10 8 3.86915646858 × 10 − 13 kshana (26)

Thus, we have n = 4.942359860 × 10²⁰ kshanas and 1 kshana = 2.0233249466 × 10⁻²¹ sec.

The Planck constant h = 6.626070040 × 10⁻³⁴ J∙sec is converted to a value that has a time unit of kshana instead of sec. Dividing the Planck constant by “n”, the new value of h will be

h ′ = h n = 6.62607004081 × 10 − 34 4.942359860 × 10 20 J ⋅ kshana (27)

Thus, the Planck constant h' = 1.34066928117 × 10⁻⁵⁴ J. Kshana.

From Equation (5), we have the velocity of light c' = π3.8615926758 × 10⁻¹³/2 = 6.0657755907 × 10⁻¹³ meters/kshana. Alternatively, we can find the velocity of light c' = c/n = 2.99792458 × 10⁸/4.942359860 × 10²⁰ = 6.0657755908 × 10⁻¹³ meters/kshana.

In the SI system, the absolute permittivity of the medium is ϵ₀ = 8.854187817 × 10⁻¹² coul²/nt∙m². The unit of ϵ₀ can be written as coul²∙sec²/kg∙m³. The value of ϵ₀ then when the time unit is kshana is ε ′ 0 = 8.854187817 × 10⁻¹² × (4.942359860 × 10²⁰)² = 2.16280546198 × 10³⁰ coul²/kg∙m³∙kshana⁻².

By the law of conservation of energy, the mass of the electron can be found as shown below.

h ′ ν ′ = m 0 c ′ 2 (28)

where h', ν' and c' are the Planck constant, gamma ray frequency and velocity of light, respectively, in which the time unit is kshana. Substituting c' = λ₁ν' meters/kshana in the above equation, we have

h ′ = m 0 λ 1 2 ν ′ 1 (29)

h ′ = m 0 λ 1 2 T ′ s (30)

where T ′ s = 1 / ν ′ is the spinning period of the electron, which is 4 kshana. Thus

h ′ = m 0 λ 1 2 4 (31)

Rearranging the terms in the Equation (31), we have

m 0 = 4 h ′ λ 1 2 (32)

Substituting the values of h' = 1.340669 × 10⁻⁵⁴ joule. kshana (Equation (27)) and λ₁ = 2.42631 × 10⁻¹² meters, we have a mass of the electron m₀ = 0.91093834243469 × 10⁻³⁰ kg = 9.1093834243469 × 10⁻³¹ kg, which is the same as the reported CODATA value (m_e = 9.10938356 × 10⁻³¹ kg) [10]. The rest mass of the electron can also be found by the law of conservation of energy when the time unit is kshana, as shown below.

m 0 c ′ 2 = 0.51 n 2 MeV (33)

where n is the number of kshanas in a second. Substituting c' = πR_s/2, and n = 1.90853806367 × 10⁸/R_s in the above equation, we have

m 0 π 2 R s 2 4 = 0.51 R s 2 ( 1.90853806367 × 10 8 ) 2 (34)

m 0 = 4 × 0.51 × 10 6 × 1.60217662208 × 10 − 19 π 2 ( 1.90853806367 × 10 8 ) 2 kg (35)

m 0 = 9.0915757 × 10 − 31   kg (36)

The calculated mass of an electron from Equation (32), which originates from wavelength λ₁ of gamma radiation is the same as the electron mass m_e = 9.10938356 × 10⁻³¹ kg as reported in CODATA [10] and it shows that the time required in pair production is about four kshanas i.e., equal to the spinning period of an electron. In a kshana electron mass formation is h ′ / λ 1 2 = 0.22773458 × 10⁻³¹ kg and in four kshanas it is equal to the reported value of electron mass (Equation (32)) [10]. It shows that the physical change in the matter i.e., electron/positron production is associated with the kshana or moment and its succession. Thus, one can know the end of its succession only at the end of the physical change [1] [2].

The generation of electron and positron is due to the electromagnetic radiation or photon. The electromagnetic wave does not disappear but gets transformed into electron and positron [15]. It relates to mass of the electron and wavelength of the gamma radiation. This is like the concept of electromagnetic origin of mass particle. Erik Haeffner (2000) proposed a concept called Condensed Electromagnetic Radiation (CER) as the electromagnetic origin of mass particles. Erik Haeffner says, “The new concept CER (Condensed Electromagnetic Radiation), proposed in this article, indicates an electromagnetic origin of mass particles, in fact, an overwhelming amount of experimental evidence confirms that the CER concept is fundamental for the physical explanation of mass particle properties [14].”

Velocity of light in meter/second is

c = 1 μ 0 ε 0 (37)

Rewriting the above equation when the velocity of light and absolute permittivity has the time unit kshana

c ′ = 1 μ 0 ε ′ 0 (38)

However, by the definition of kshana, the velocity of light c' = πR_s/2 m/kshana and ε ′ 0 = ε 0 × n 2 . Therefore, from Equation (38), we have

n = 2 c π R s (39)

The Equation (39) is the same as the Equation (6) or Equation (8).

When the incident photon energy is such that it only liberates an electron from the metal surface without any kinetic energy (i.e., mv²/2 = 0). In such case hν₀ = W_ϕ. where ν₀ is the threshold frequency in cycles per second and W_ϕ is the work function in joule. Now rewriting the equation for the work function having time unit kshana, we have

h ′ ν ′ 0 = h ′ c ′ λ ′ 0 = w ′ ϕ (40)

For h' = h/n, W ′ ϕ = W ϕ / n 2 , and λ 0 = λ ′ 0 = c ′ / ν ′ 0 meters, we have

n = λ ′ 0 w ϕ h c ′ (41)

n = 2 λ ′ 0 w ϕ h π R s since c ′ = π R s 2 (42)

For a tungsten cathode of threshold wavelength λ ′ 0 = λ₀ = 2300 × 10⁻¹⁰ m, and work function W_ϕ = 5.38 eV [16], we have

n = 1.9085380 × 10 8 R s kshana / sec (43)

The above equation is same as Equation (9).

Alternatively, from Equation (40), we can find the value of n as shown below:

h ′ = w ϕ n 2 ν ′ 0 = w ϕ n 2 ν 0 / n = w ϕ n ν 0 (44)

Since W ′ ϕ = W ϕ / n 2 , ν ′ 0 = ν 0 / n , where n is the number of kshanas in one second. From Equation (31), h ′ = m 0 λ 1 2 / 4 . Therefore,

w ϕ n ν 0 = m 0 λ 1 2 4 (45)

n = 4 w ϕ m 0 λ 1 2 ν 0 = 4 w ϕ λ 0 m 0 λ 1 2 c (46)

1) For a tungsten cathode with a threshold wavelength λ₀ = 2300 × 10⁻¹⁰ m, work function W_ϕ = 5.38 eV [16], and electron wavelength λ₁ = 2.42631 × 10⁻¹² (Section 5.1), From Equation (46), the value of n will be

n = 4.932626423359 × 10 20     kshana (47)

2) For Aluminum work function is 4.25 eV [16]. The threshold frequency will be

ν 0 = 1.0276454364795 × 10 15   Hz (48)

and threshold wavelength λ₀ = c/ν₀ = 2.91727523 × 10⁻⁷ m. Thus, from Equation (46)

n = 4.94236082 × 10 20     kshana (49)

3) For Rb, the work function is 2.16 eV [16]. Then, the threshold frequency will be ν₀ = 0.5222856806 × 10¹⁵ Hz, and the threshold wavelength will be λ₀ = c/ν = 5.740009 × 10⁻⁷ m. Thus,

n = 4 w ϕ m 0 λ 1 2 ν 0 (50)

n = 4.9423608678 × 10 20     kshana (51)

4) For Mg, the work function is 3.66 eV [16], and the threshold frequency will be ν = 0.88498407 × 10¹⁵ Hz.

n = 4 w ϕ m 0 λ 1 2 ν 0 (52)

where, λ₁ = 2.42631 × 10⁻¹² meters.

n = 4.94236089 × 10 20     kshana (53)

The number of kshanas in a second can also be found by taking the ionization energy of the hydrogen atom, which is 13.6 eV. This is the energy needed to free the electron from the nucleus of the hydrogen atom. In the Lyman series for n = 1, the associated wavelength is 1026 × 10⁻¹⁰ meters, and the energy difference is −3.4 − (−13.6) = 10.2 eV [9]. Now we can estimate the value of n as shown below. By the law of conservation of energy, we have

h ′ ν 0 = 10.2 × 1.6021766208 × 10 − 19 n 2 (54)

h ′ ν 0 n = 10.2 × 1.6021766208 × 10 − 19 n 2 (55)

where ν 0 = n ν ′ 0 = c / λ 0 and λ₀ = 1026 × 10⁻¹⁰ meters is the wavelength of the first member of the Lyman series. Thus,

h ′ = 10.2 × 1.6021766208 × 10 − 19 n ν 0 (56)

h ' = 10.2 × 1.6021766208 × 10 − 19 λ 0 n c (57)

h ' = 6628.6247 × 10 − 37 n = m 0 λ 1 2 4 (58)

n = 4 × 6628.6247 × 10 − 37 m 0 λ 1 2 (59)

Substituting the values of electron rest mass m₀ and λ₁, we get

n = 4.944266452 × 10 20     kshana (60)

Equations (26), (47), (48), (49), (51), (53), and (60) show that the number of kshanas in a second is the same, i.e., n = 4.942359860 × 10²⁰ kshanas and 1 kshana = 2.0233249466 × 10⁻²¹ sec.

The statistical concept of movement of electron can be considered for validation of kshana. For a free spinning electron with effective Lande’-g factor g* = 2, the radius R_s of spinning electron is [8]

R s = 5 g * ℏ 4 m 0 c = 5 g * ℏ 8 π m 0 c (61)

where Planck constant h = 6.626070040 × 10⁻³⁴; J∙s, rest mass of free electron m₀ = 9.10938356 × 10⁻³¹ kg, and velocity of light c = 2.99792458 × 10⁸ m/s. Substituting these values, we get

R s = 9.6539816887 × 10 − 13     meter (62)

Converting Equation (61) for radius with time unit kshana, we get

R s = 5 g * h ′ 8 π m 0 c ′ (63)

where Planck constant h' = 1.34066928117 × 10⁻⁵⁴ J. kshana, rest mass of free electron m₀ = 9.10938356 × 10⁻³¹ kg, and velocity of light c = 6.0657755908 × 10⁻¹³ meters/kshana. Substituting these values, we get

R s = 9.653981690 × 10 − 13     meter (64)

Equations (62) and (64) give the same result even though time units of Planck constant h and velocity light c are different.

Spinning period for free electron is (Ziya Saglam et al. [8])

T s = 8 π R s 2 m 0 5 g * ℏ = 2.023324947 × 10 − 20 sec (65)

Rewriting the Equation (65) in which h is replaced with nh'/2π. Since h = h/2π and h = nh' where n is the number of kshanas in a second and h' is in J∙kshana or kg∙m²/kshana. Thus, the number of kshanas in a second can be calculated using the rewritten formula for T_s

T s = 8 × 2 π 2 R s 2 m 0 5 g * n h ′ (66)

n = 16 π 2 R s 2 m 0 5 g ∗ h ′ T s kshanas / sec (67)

For effective g-factor g* = 2,

n = 4.9423598634677 × 10 20 kshanas / sec (68)

Thus, the value of “n” in Equation (68) is same as in Equations (26), (47), (48), (49), (51), (53), (60) and show that the number of kshanas in a second is the same, i.e., n = 4.942359860 × 10²⁰ kshanas and 1 kshana = 2.0233249466 × 10⁻²¹ sec.

Alternatively, the ultimate indivisible value of the kshana which is 2.02332494691 × 10⁻²¹ sec is in good agreement with Ziya Saglam et al. [8]. Ziya Saglam et al. calculated the spinning period for a free electron which is 1.9 × 10⁻²⁰ sec (for effective Lande’-g factor, g* = 2). When this value of period is divided by 4 kshanas (since spinning period of free electron is 4 kshanas) give the value of 1 kshana equal to 4.75 × 10⁻²¹ sec for free electron. Both the values are of the same order of magnitude i.e., 10⁻²¹ sec. Ziya Saglam also calculated the period of spin in an atomic state which is T_{s (n = 1, l = 0, mj = 0, ms = 1/2)} = 1.48 × 10⁻²¹ sec (for effective Lande’-g factor, g* = 1). Again, this value of period is divided by 4 kshanas give the value of 1 kshana equal to 0.37 × 10⁻²¹ sec for period of spin in an atomic state which also has same order of magnitude i.e., 10⁻²¹ sec.

The value of a kshana can also be determined using the circular frequency of a spinning electron given by Olszewski [17], which is 2π/T₂ = m_ec²/h = 0.78 × 10²¹ sec⁻¹ [17]. Where, period T₂ = 2π/0.78 × 10²¹ sec⁻¹ = 8.055365778 × 10⁻²¹ sec and a kshana is T₂/4 = 8.055365778 × 10⁻²¹ sec/4 = 2.0138414446 × 10⁻²¹ sec. Again, it is same as shown in the above sections of this paper.

In a Compton scattering, the Compton wavelength λ_C = h(1 − cosθ)/m₀c = h/m₀c, whose value is 2.4263102367 × 10⁻¹² meters [10] for θ = π/2, where h the plank constant, m₀ is the mass of electron, c is the velocity of light and θ is the angle of scattering. Now, the Compton frequency ν_C = c/λ_C = 2.99792458 × 10⁸/2.4263102367 × 10⁻¹² = 1.2355899 × 10²⁰ Hz (or ν_C = ω₀/2π = mc²/2πh [18]), and period will be T_C = 1/ν_C = 0.80932997877 × 10⁻²⁰ sec. Therefore, 1 kshana = T_C/4 = 0.80932997877 × 10⁻²⁰/4 = 0.20233249 × 10⁻²⁰ sec i.e., 2.0233249 × 10⁻²¹ sec and n = 4.942359859269 × 10²⁰ kshanas, which is same as given in the above sections.

Again, it shows that the value of “n” is same as in Equations (26), (47), (48), (49), (51), (53), (60) and show that the number of kshanas in a second is the same, i.e., n = 4.942359860 × 10²⁰ kshanas and 1 kshana = 2.0233249466 × 10⁻²¹ sec.

The Compton wavelength is given by the equation λ_C =h/m₀c whose value is 2.4263102367 × 10⁻¹² m. The reduced Compton wavelength is Λ_C = λ_C/2π = h/2πm₀c which is equal to 3.8615926764(18) × 10⁻¹³ m [10]. Writing the reduced Compton wavelength equation where the time unit is kshana is as shown below

Λ C = λ C 2 π = h ′ 2 π m 0 c ′ meter (69)

Substituting the Planck constant h' = 1.34066928117 × 10⁻⁵⁴ J. kshana, the velocity of light c' = 6.0657755908 × 10⁻¹³ meters/kshana, and mass of the electron m₀ = 9.10938356 × 10⁻³¹ kg, we obtain a reduced Compton wavelength equal to 3.8615926760 × 10⁻¹³ m which is the same as the reported CODATA value.

The orbital velocity in the first orbit of the hydrogen atom is v' = 2.18769126277 × 10⁶/4.942359860 × 10²⁰ = 4.426410307 × 10⁻¹⁵ meters/kshana. Fine structure constant is given by the following equation

α = v ′ c ′ = 4.426410307 × 10 − 15 6.0657755908 × 10 − 13 (70)

or

α = 0.7297352565620 × 10 − 2 = 1 137.0359991528 (71)

Thus, the fine structure constant is the same as that reported [10].

Using the following equation [9], the Rydberg constant R_∞ is found as shown below.

R ∞ = m 0 e 4 8 ε 0 2 h ′ 3 c ′ (72)

R ∞ = 60.0247762251 × 10 − 107 546.9856635635 × 10 − 115 / m (73)

The value of the Rydberg constant R_∞ is equal to 10973738.4768/meter. This agrees with the reported value R_∞ = 10973731.568525(73)/m [10].

The spin angular momentum J is given by the following Equation [7], where the time unit is in seconds.

J = m 0 R s 2 w 2 = m 0 R s 2 w ′ 2 (74)

where ω is rad/sec and ω' = π/2 rad/kshana are the angular frequencies. From equation c' = πR_s/2 m/kshana [2], we can write the above equation as

J ′ = m 0 R s c ′ 2 (75)

In semi-classical model of spinning electrons, it is assumed that [7].

R s = Λ c = ℏ m 0 c = h 2 π m 0 c = h ′ 2 π m 0 c ′ meter (76)

Substituting the value of R_s from Equation (76) in Equation (75) we have

J ′ = h ′ 4 π = 1.3406690 × 10 − 54 4 π (77)

J ′ = 0.106687049 × 10 − 54 (78)

Converting time unit kshana to time unit sec, we have

J = 0.106687049 × 10 − 54 × 4.942359860 × 10 20 (79)

Thus, spin angular momentum is equal to J = 0.527285789548 × 10⁻³⁴ which is in good agreement with the value provided by CODATA [10].

The spin magnetic moment of a simple model of spinning electrons [7], is

μ ′ = I S = ( e T ′ s ) ( π R s 2 ) = ( e 4 ) ( π R s 2 ) (80)

where current I = e / T ′ s and e is the electronic charge. For a simple model of the spin magnetic moment S = π R s 2 [7], T ′ s = 4 kshanas is the spinning period of the electron. From Equation (80)

μ ′ = ( e 4 ) ( π R s h ′ 2 π m 0 c ′ ) (81)

μ ′ = ( R s 4 c ′ e h ′ 2 π m 0 ) (82)

From Equation (11) keeping the value of c', we have

μ ′ = ( e h ′ 4 π m 0 ) = 0.018764331838931 × 10 − 42 (83)

where h' has the unit, Joule∙kshana. When it is converted to the time unit sec, we have a Bohr magneton value of 0.092740080 × 10⁻²² = 9.27400 × 10⁻²⁴ J∙T⁻¹ which is equal to the reported value of Bohr magneton µ_B = 927.4009994(57) × 10⁻²⁶ J∙T⁻¹.