_{1}

^{*}

The research work is carried out to find spectrum of wavelength of the emitted radiations from filament bulb. Both temperature and wavelength measurement are based on theoretical model. The temperature finding formula for tungsten filament is obtained by using blackbody radiation approach. The peak wavelength of the emitted radiation is obtained 1.461 μm and 1.125 μm for 6 and 500 watt bulb respectively by using Wein’s displacement law which depends upon temperature of the filament. The wavelength obtained by using Wein’s displacement law, is just an index, which helps to investigate, “how the radiation energy density is distributed” so as to give rise to an electromagnetic spectrum. The results obtained from the applied methodologies show that the accuracy of a model is quite good. Some mathematical techniques and probability theory are used to verify the work. The work is followed by both classical and quantum analysis to justify the results. Temperature only is not the key factor that deserves power of the bulb. The value of resistance plays a vital role in fixing power of the bulb. At least one factor is important in the calibration of the power of the bulb, “either temperature or surface area” of the filament.

Tungsten is a chemical element with the symbol W. Its atomic number is 74. Tungsten has melting point of 3422˚C and boiling point is 5930˚C. Its density is 19.3 gm/cm^{3} at room temperature and 17.6 gm/cm^{3} when liquid is at melting point. Its heat of fusion is 52.31 kJ/mol and heat of vaporization is 774 kJ/mol. The value of molar heat capacity of tungsten is 24.27 J/(mol∙K). Its density is about 1.7 times than that of lead. It lies in the d-block of the periodic table. Its group and period in the periodic table is “6”. Tungsten is named by Torbern Bergman in 1781. It is paramagnetic substance [

Its phase at STP is solid. Pure single crystalline tungsten is more ductile. The strategic value of tungsten came to notice in the early 20^{th} century. Tungsten desirable properties such as resistance to high temperatures, its hardness and density, and its strengthening of alloys made it an important raw material for the arms industry [

Tungsten is a metal which presents many points of particular interest both from the practical and the scientific point of view. From the time when tungstic acid was first prepared by Scheele in 1781 and Bergmann separated the metal, tungsten remained a rare metal, and it only began to assume industrial importance as the result of the work of Oxland in 1847-57 [

A great deal of research has been undertaken to investigate the optical, electrical, chemical and thermal properties of tungsten materials; as well as the characteristics of tungsten light bulbs [

greater pressure difference than that of the outer air. The incandescent lamp is widely used in household and commercial lighting, for portable lighting such as table lamps, car headlamps, flashlights, and for decorative and advertising lighting [

The temperature of the filament increases due to Joule heating on biasing electrical current. An inert gas like argon is normally used to fill inside the bulb which prevents the tungsten filament from oxidizing, otherwise results in the catastrophic failure of the transducer [_{Wire} will depend on the operating temperature T of the filament [

The radiation in the filament bulb is emitted by a process called incandescence. The electrical resistive heating creates thermally excited atoms. Some of the thermal kinetic energy is transferred to electronic excitations within the solid. The photonic emission occurs when the atoms from the excited states jump to the lower state [

A hot metal filament glows red, and when heating continues, its glow eventually covers the entire visible portion of the electromagnetic spectrum. The temperature (T) of the object that emits radiation, determines the wavelength at which the radiated energy is at its maximum [

There are no sharp, well defined boundaries in the electromagnetic spectrum. Conventionally; the optical radiation range is defined as extending from 1 mm at the bottom end of the infrared to 100 nm at the upper end of the ultraviolet. In

The colour of the radiations depends upon the wavelength of the emitted radiations. Emitted wavelength and temperature of the heating body are inversely proportional to each other [

The filament lamp will shine a reddish light at 2000 K because it emits the visible light of longer wavelength. At 3000 K, the filaments shine brighter and emit a yellow light of shorter wavelength [

Stefan-Boltzmann and Wein’s displacement laws explain respectively, the intensity and color of light emitted by a body [

Visible light is a form of electromagnetic radiation which can be perceived by our eyes. Different visible colours of light with their wavelength limit have been shown in

Radiations | Wavelength range |
---|---|

Ultraviolet | 100 - 400 nm |

Light | 380 - 400 to 760 - 780 nm |

Infrared | 760 - 780 nm to 1 mm |

Color | Wavelength (nm) |
---|---|

Red | 625 - 740 |

Orange | 590 - 625 |

Yellow | 565 - 590 |

Green | 520 - 565 |

Cyan | 500 - 520 |

Blue | 435 - 500 |

Violet | 380 - 435 |

nm to 740 nm. Various spectral color bands make up light. The band from 100 nm to 400 nm is called ultraviolet radiation. The band from 710 nm to 1.5 micrometers is called near infrared and the band from 1.5 to 4.0 micrometers is called far infrared [

The monochromatic radiant emittance and wavelength λ are related by the conventional spectrum curve of the blackbody thermal radiations [

Planck found an empirical formula to explain the experimentally observed distribution of energy in the spectrum of a black-body [

Planck’s formula is given by

E d λ = 8 π h c λ 5 ⋅ 1 e h c λ k T − 1 d λ (1)

For Shorter wavelength e h c λ k T becomes large compared to unity and hence the Planck’s law reduces to

E d λ = 8 π h c λ 5 ⋅ 1 e h c λ k T d λ = 8 π h c λ 5 ⋅ e − h c λ k T d λ

This is Wein’s law [

For longer wavelength e h c λ k T may be approximated to ( 1 + h c λ k T ) and hence the Planck’s law reduces to

E d λ = 8 π h c λ 5 1 1 + h c λ k T − 1 d λ = 8 π h c λ 5 1 h c λ k T d λ = 8 π h c λ 5 λ k T h c d λ = 8 π k T λ 5 d λ (2)

This is Rayleigh’s Jeans law [

Thus we see that Wein’s displacement law holds for shorter wavelengths while the Rayleigh’s Jein’s law holds for longer wavelength [

For a temperature ranging from room to about 2500 K, tungsten filament obeys a linear relation between its resistance and temperature,

R ( T ) = R 0 [ 1 + α ( T − T 0 ) ] (3)

Where R 0 represents ambient resistance measured at temperature T 0 and α is temperature coefficient of resistivity. R ( T ) is the resistance at temperature T [

Electromagnetic radiation is a form of energy that is all around us and takes many forms, such as radio waves, microwaves, x-rays and gamma rays. The electromagnetic waves are divided into different ranges, depending on wavelength and corresponding frequency [

The emissivity of tungsten wire filaments in incandescent lamps change with wavelength. Tungsten wire is the key component of the incandescent lamp. The lamp shows better performance if the emissivity of the filament is higher [

The energy emitted per unit area per unit time depends upon the emissivity of the material. The total emissivity can be calculated by using the relation as below:

P = ε σ A T 4 (4)

where ε is the total emissivity, σ is Stefan’s Constant, A is the area of the tungsten, T is the temperature [

The bulb filament is not efficient at emitting blackbody radiation. The emissivity is defined as the ratio of energy radiated by a material to that radiated by

Temperature Degrees K (T) | Spectral Emissivity (ε) |
---|---|

2100 | 0.433 |

2200 | 0.431 |

2300 | 0.429 |

2400 | 0.427 |

2500 | 0.425 |

2600 | 0.423 |

2700 | 0.421 |

2800 | 0.419 |

2900 | 0.417 |

3000 | 0.415 |

3100 | 0.413 |

an ideal blackbody at the same wavelength and temperature. The emissivity and temperature of the filament has the relation of 1 ε filament ( T filament ) [

The performance of lamp depends upon the emissivity of filament. There is direct correlation between emissivity and illumination. The life of the filament depends strongly on the burning temperature. The key in measuring the filament temperature is the emissivity [

The Stefan-Boltzmann constant σ has been evaluated by using an absolute radiometer of the electrical substitution type to measure the radiance of a cavity radiator at the freezing point of gold. The value obtained for σ is (5.6644 ± 0.0075) × 10^{−8} W∙m^{−2}∙K^{−4} [

An atom may be excited to a higher level due to the presence of sufficient energy. If this excited atom transits back to its ground state, this energy is then released at a characteristic frequency related to these two energy levels as shown in _{0} is the energy of the ground state and E_{1} is the energy of the first excited state then the atom in the first excited state releases energy when drops to the ground state. The released energy is equal to the difference in energy between these two levels [

Temperature (K) | Resistivity (μΩ∙cm) |
---|---|

2100 | 60.06 |

2200 | 63.48 |

2300 | 66.91 |

2400 | 70.39 |

2500 | 73.91 |

2600 | 77.49 |

2700 | 81.04 |

2800 | 84.70 |

2900 | 88.33 |

3000 | 92.04 |

E = h f = h c λ = E 1 − E 0 (5)

where h is the Planck’s constant, f is the frequency, c is the velocity of light and λ is the wavelength of the emitted radiation.

A blackbody can be described as an enclosed body which absorbs all the incident radiations upon it. When the blackbody is at equilibrium, it will emit radiation at the same rate as it absorbs radiation from the surrounding medium. Sometimes a blackbody is referred to as a complete absorber [

At a particular wavelength λ , the radiation emitted by a perfect blackbody radiator is described by Planck radiation law

I ( λ , T ) = 2 π h c 2 λ 5 1 e − h c λ k T − 1 [

Yang X., Wei, B. studied the local spectrum characteristics of the blackbody thermal radiation based on the traditional spectrum curve.

The wavelength of maximum emission of any body is inversely proportional to its absolute temperature. In this way, higher the temperature, the wavelength

of maximum emission is shorter. This condition is called Wein’s law. Wein’s law tells that if the temperature of a body increases, the wavelength of maximum emission becomes smaller [

The following equations describe this law;

λ max = b T (7)

where b is a constant equal to 2897 μm and T is the temperature in Kelvin.

Nine theoretical filament bulbs have been taken in this research work. These bulbs are chosen considering, they are the more frequently used bulbs in our daily life. The filament in the tungsten lamps has been assumed to be essentially a single-coil straight wire. The parameters length L, emissivity ε , and resistivity ρ are the functions of temperature shown in Equation (14). The 60 and 100 W lamps are basically coiled coil types [

There are different techniques to determine the temperature of the filament bulb. In this work, the power of the filament bulb has been calculated in terms of voltage, radius of the filament wire, its length and resistivity using Equation (12). The relation between power and intensity is shown in

radiation curve. Microsoft Power Excel 2010, and python programming has been used to plot the graphs. The different graphs at different temperatures, between spectral emittance and wavelength are plotted with the help of high resolution blackbody radiation calculator [

radiant emittance, peak spectral radiance, band radiance, and wavelength of peak are calculated by using blackbody spectral calculator [^{−7} m and 6 × 10^{−6} m respectively. The model of choosing lower value and upper value of wavelength is based on the probability distribution (probability occurrence) of the spectrum of emitted radiations from filament bulb. The graph generated by using Hi-resolution spectral calculator [

Mean value, standard deviation, minimum value, maximum value, difference between extreme values of wavelength, difference between extreme values of spectral emittance, have been calculated using vernier spectral analysis software [

Provided whole digits after decimal (of length and diameter) are included into research work. The provided value of length of filament used is in three decimal places, diameter of filament is in 6 decimal places. Calculated radius, from the

provided value of diameter, has been kept in seven decimal places. The purpose of including almost all digits after decimal, of these physical quantities, is to minimize the possible error that may occur during calculation. Scientific calculator has been used to calculate radius of tungsten filament, current flowing in filament, power dissipated in filament, and surface area of filament. Normal distribution application is used to plot the distribution curve.

When a light bulb is turned on using a switch, a constant (ac) voltage V of 120 V is applied across the filament. Since the filament has a high resistance, because of its fine diameter and long length (see Equation (11)), a small amount of current flows through the filament according to Ohm’s law

V = I R (8)

where I is current, V is voltage across the filament and R is resistance of the filament [

P = V 2 R (9)

Since voltage is a constant in our electrical distribution system, the form of equation for the power dissipated in a resistor should be that shown in Equation (9), and should not be P = I 2 R . Since V is a constant, Equation (9) relates one variable R to another variable P. If the equation P = I 2 R was used, then P, I and R would all be variables. The filament is compared to a theoretical blackbody radiator. The total power emitted per unit surface area (A) of a hot object at temperature T (in Kelvin) is given by the Stefan-Boltzmann law [

P A = ε σ T 4 (10)

where, σ is called Stefan-Boltzmann constant, and has the value 5.67 × 10^{−8} W/(m^{2}-K^{4}). The emissivity ε is a material dependent quantity. For tungsten, value of ε is taken as 0.421 assuming that the temperature of filament reaches to about 2700 K [

R = ρ L A (11)

Combining Equation (9) and Equation (11), the power dissipated in the filament can be written as

P = V 2 π r 2 ρ L (12)

Since voltage in our houses is fixed, it is apparent from Equation (9) that for higher wattage bulbs, electrical resistance of the filament must decrease as bulb wattage increases. From Equation (11), R can be decreased by increasing A (i.e. r^{2}) or by decreasing L. Therefore, for higher wattage bulbs, it is necessary to either increase r or decrease L. From the Stefan-Boltzmann law [

P = ε σ T 4 ( 2 π r L ) (13)

From Equation (13), we can see that the bulb wattage (brightness), filament temperature (life time), filament radius and length are all interdependent. The surface area of the filament can be maximized by increasing r. From Equation (13), if the power is increased, it is desirable to increase r and L to minimize increase in temperature T. Equation (12) and Equation (13) can be equated and we get

T 4 = V 2 r 2 ε σ ρ L 2 (14)

The Equation (14) shows interdependence of filament temperature, radius and length. Tungsten filament emissivity directly affects relationship between surface temperature of a filament and its thermal radiation spectrum.

All digits after decimal are taken, to calculate the surface area of the filament. The obtained digits after decimal are 10 in 6, 10, 25, 40, 60, 100, 200, 300 watt bulb and nine digits in 500 watt bulb. So, the error in the calculation is expected to minimum. The temperature of glass that surrounds the filament is taken to be very small in compared to the temperature of the tungsten filament. So the energy emitted per unit area per unit time (Intensity of the radiations) is supposed to be dependent upon the temperature of the tungsten filament.

There is no any theoretically derived direct equation, to calculate the wavelength of the emitted radiations from the filament bulb. So since the filament bulb acts as a source of black body, Wein’s displacement law in Equation (15) is used to calculate the peak wavelength of the emitted radiation. Although, Wein’s displacement law in Equation (15) is used only for higher energy and shorter wavelengths, the law is used, considering the spectrum of the radiations from the filament bulb that may consist of very few fraction of visible light at the shorter wavelengths. The ultraviolet radiation (320 - 340 nm), lie in shorter wavelength [

Lamp watt P | Length L cm | Diameter D cm |
---|---|---|

6 | 37.084 | 0.001143 |

10 | 43.180 | 0.001626 |

25 | 56.388 | 0.003048 |

40 | 38.100 | 0.003302 |

60 | 53.340 | 0.004572 |

100 | 57.912 | 0.006350 |

200 | 63.500 | 0.009652 |

300 | 72.390 | 0.012700 |

500 | 87.376 | 0.018034 |

The wavelength of the emitted radiation has been calculated by using the following relation.

Wavelength ( λ ) = h c 4.965 k T (15)

where h is the Planck’s constant, c is the velocity of light, k is the Boltzmann’s constant, and T is the temperature of the body.

There is a distribution of different photons of various different energies. The peak of the distribution tells us that which of these photons occurs at the greatest amount at the given temperature. The traditional spectrum curve in ^{−6}. This means maximum of the emitted radiations lie in the near infrared region. The work is also based on the model that, “Probability occurrence of the visible light is the quantum result”.

The work is based on the fact that, “No radiations have actual value of wavelength”. They are all approximate. There is no any strong evidence that could be justified by theoretical and experimental procedure on the validity of the provided value of the wavelength which we are using on these days. Thus the various values of the wavelength (which I have mentioned above in the introduction and in the literature review) fall within certain particular range are only probabilistic.

To find the percentage of the wavelength range of the emitted radiations from the filament bulb, the total number of square divisions on the graph was counted. The total numbers of square divisions are supposed as 100%. Since the curve does not occupy exactly integral square division; Fractional divisions are also taken into calculation. Calculation was done manually. Even, one fifth of a square division has been included in counting.

The value of emissivity is taken as 0.421 assuming the filament temperature reaches to about 2700 K. The value of resistivity is taken to be about 8.104 × 10^{−7} ohm-m. The current passed depends upon the resistance of the filament. The resistance can be varied by varying the length and diameter of the filament. The only constant parameter is applied voltage.

The temperature of the filament depends upon the applied voltage, emissivity, radius, length, resistivity of the filament, and the value of Stefan’s constant. The temperature of the filament, when the bulb is on, is calculated by using Equation (14). The supplied voltage is constant. In

Length L cm | Diameter D cm | Radius r cm | Temperature in Kelvin (K) T 4 = V 2 r 2 ε σ ρ L 2 | Bulb Power P watt |
---|---|---|---|---|

37.084 | 0.001143 | 0.0005715 | 1983 | 6 |

43.180 | 0.001626 | 0.0008130 | 2007 | 10 |

56.388 | 0.003048 | 0.0015240 | 2055 | 25 |

38.100 | 0.003302 | 0.0016510 | 2551 | 40 |

53.340 | 0.004572 | 0.0022860 | 2338 | 60 |

57.912 | 0.006350 | 0.0031750 | 2436 | 100 |

63.500 | 0.009652 | 0.0048260 | 2583 | 200 |

72.390 | 0.012700 | 0.0063500 | 2591 | 300 |

87.376 | 0.018034 | 0.0090170 | 2575 | 500 |

The temperature of the heating filament depends upon its length and the diameter. So the power of the bulb depends upon the length and diameter of the filament. The volume of the material of the filament and the surface area of the filament increases, if the power of the bulb increases, except for 25 watt and 40 watt bulb. The volume and surface area of 25 watt bulb is greater than the 40 watt bulb. But the interesting factor is that the temperature of 40 watt bulb is greater than the 25 watt bulb. Thus in the above result from

At constant voltage, the value of resistance depends upon the power of the bulb. So the current flowing through the filament varies with resistance. The power of the bulb in column (5) of

The resistance is maximum in the lower wattage bulb and minimum in the higher wattage bulb which can be observed in

The calculated values of energy emitted per unit area per unit time (Intensity) from each wattage bulb are shown in

Bulb Wattage (P) | Resistance R = V 2 P | Operating voltage (V) | Current (I) | Power dissipated in the filament (P) = I^{2}R | Area of the tungsten filament (A) = 2πrl in m^{2} |
---|---|---|---|---|---|

6 | 2400 | 120 | 0.05 | 6 | 0.0000133095 |

10 | 1440 | 120 | 0.0833 | 10 | 0.0000220462 |

25 | 576 | 120 | 0.20833 | 25 | 0.0000539674 |

40 | 360 | 120 | 0.33333 | 40 | 0.0000395031 |

60 | 240 | 120 | 0.5 | 60 | 0.0000765753 |

100 | 144 | 120 | 0.8333 | 100 | 0.0001154707 |

200 | 72 | 120 | 1.67 | 200 | 0.0001924512 |

300 | 48 | 120 | 2.5 | 300 | 0.0002886768 |

500 | 29 | 120 | 4.17 | 500 | 0.0004947820 |

Power (P) in watt | Intensity (I) in W/m^{2 } |
---|---|

6 | 369,110 |

10 | 387,306 |

25 | 425,709 |

40 | 1,010,897 |

60 | 713,252 |

100 | 840,571 |

200 | 1,062,583 |

300 500 | 1,075,808 1,049,480 |

The result of the intensity for each wattage bulb, shown here in

The illuminating areas of the surface from which the radiations are emitted depend upon the length and diameter of the tungsten filament of the incandescent bulb. Here in ^{2} (see ^{2} (see

The resistance and current variation is shown in

Lower the value of filament resistance, greater is the flow of electric current. The higher wattage bulb has lower resistance filament and maximum current flow. The fall in resistance with respect to the lower wattage bulb is greater than the fall in resistance for higher wattage bulb. The change between any two resistances is greater for lower wattage bulbs than higher wattage bulbs as shown in

Δ R Δ P , cannot be zero, because Δ R cannot be made zero. This means that power of the filament bulb cannot be increased by lowering the resistance after certain limit.

The value of emissivity does not largely affect the value of the temperature. The Wein’s displacement law is considered to verify the statement and is preceded as below:

Wavelength ( λ ) = h c 4.965 k T

For ε = 0.421 , T = 1983 K

Wavelength ( λ ) = 6.62 × 10 − 34 × 3 × 10 8 4.965 × 1.38 × 10 − 23 × 1983 = 1.46 × 10 − 6 m

Similarly, when ε = 0.386 , (which is effective emissivity and is equal to the 90% of the 0.421), the value of temperature found to be 2027 K. So, on using

Wavelength ( λ ) = 6.62 × 10 − 34 × 3 × 10 8 4.965 × 1.38 × 10 − 23 × 2027 = 1.43 × 10 − 6 m .

The wavelength, on two different values of emissivity is not largely affected. The percentage error in the two values of the temperatures with different emissivity (% error) = 2.2 {the two temperatures taken here for the calculation are 1983 K and 2027 K}.

The value of the emissivity lies within the range of 0.413 to 0.470, depending on the temperature of the heating filament [given in

The temperature and peak emission wavelength relation is illustrated in

It can be observed from ^{−6} m. From this point of view, it can be estimated that the emitted radiations from the filament bulb carries few visible radiations and most infrared radiations. Shorter wavelengths are emitted when the temperature of the heated filament goes on increasing. The peak emission wavelength is greatest for 6 watt bulb heated at temperature 1983 K which is shown in

In this section, the bulb power is discussed on the basis of surface area, and volume of the material of tungsten filament. The surface area of the filament affects its temperature.

The relation between the power of the bulb and the surface area of the filament is illustrated in

Higher the power of the bulb, higher is the surface area of the filament as calculated in

Temperature (T) in Kelvin | Peak emission wavelength λ max = h c 4.965 k T (in meter) |
---|---|

1983 | 1.461 × 10^{−6} |

2007 | 1.444 × 10^{−6} |

2055 | 1.410 × 10^{−6} |

2551 | 1.136 × 10^{−6 } |

2338 | 1.239 × 10^{−6 } |

2436 | 1.189 × 10^{−6} |

2583 | 1.122 × 10^{−6} |

2591 | 1.118 × 10^{−6 } |

2575 | 1.125 × 10^{−6} |

Power dissipated in the filament (P) = I^{2}R (in watt) | Volume of the material of the filament (v) = πr^{2}l (in m^{3}) | Surface area of the tungsten filament (A) = 2πrl (in m^{2}) |
---|---|---|

6 | 3.8032 × 10^{−11} | 0.0000133095 |

10 | 8.9620 × 10^{−11} | 0.0000220462 |

25 | 4.0000 × 10^{−10 } | 0.0000539674 |

40 | 3.0000 × 10^{−10} | 0.0000395031 |

60 | 9.0000 × 10^{−10} | 0.0000765753 |

100 | 1.8000 × 10^{−9} | 0.0001154707 |

200 | 4.6000 × 10^{−9} | 0.0001924512 |

300 | 9.2000 × 10^{−9} | 0.0002886768 |

500 | 2.2300 × 10^{−8} | 0.0004947820 |

In this section I have discussed the behavior of the radiations emitted from the filament bulb on the basis of the parameters: wavelength and spectral emittance. The value of spectral emittance of radiations for different value of wavelength can be calculated by taking the ratio of the intensity of the emitted radiations (at a particular temperature) and wavelength. The plot between spectral emittance and wavelength in

^{11} W/m^{3}, and 84 × 10^{11} W/m^{3}.

The spectral emittance and wavelength plot fits more accurately in power curve. The power curve equation is satisfied by the form y = a x b . The root mean square error in power curve fitting is 0.002996. The value of power b in x^{b}, is −0.9999, which is constant. The value of co-efficient a is 8397.

Statistics for the plot between spectral emittance, and wavelength, in power curve fitting

x-Range = 100 - 2500 nm

y = a x b

a = 8397

b = −0.9999

RMSE = 0.002996

Standard deviation = 21.959

Mean = 14.632 × 10^{11} W/m^{3}

Minimum = 3.360 × 10^{11} W/m^{3}@2500.000 nm

Maximum = 84.000 × 10^{11} W/m^{3}@100 nm

Wavelength (nm) | Spectral Emittance (W/m^{3}) |
---|---|

100 | 8.4 × 10^{12} |

200 | 4.2 × 10^{12 } |

300 | 2.8 × 10^{12} |

400 | 2.1 × 10^{12 } |

500 | 1.68 × 10^{12} |

700 | 1.2 × 10^{12} |

1000 | 8.4 × 10^{11} |

2000 | 4.2 × 10^{11 } |

3000 | 2.8 × 10^{11} |

4000 | 2.1 × 10^{11 } |

5000 | 1.68 × 10^{11} |

The root mean square error is less in power curve fitting. The power curve equation is y = a x b . The spectral emittance and wavelength curve is satisfied by the power curve equation.

The above curve in

Here I have discussed the blackbody radiation curve for 6, 100, 500 watt bulb. The value of radiant emittance, radiance, peak spectral radiance, wavelength of peak, spectral radiance, band radiance, are calculated for these bulbs. These values show the probability distribution of the radiation energy density. The value of lower limit and upper limit of wavelength are taken respectively 0.27 μm and 6 μm respectively. These values are chosen because they contain all possible wavelengths (both shorter and longer) that may emit from the bulb.

The results obtained in this section are for 6 watt bulb. The 6 watt bulb filament gets heated at temperature 1923 K. The value of peak wavelength obtained is 1.4613 μm. The value of radiance per m^{2} per steradian is 117,502 watt. About 7.35485e+18 photons are required for one joule of energy in 6 watt bulb.

In 6 wattage bulb, no radiations exist 0.5 μm below which can be concluded from

The results obtained in this section are for 100 watt bulb. The 100 watt bulb filament gets heated at temperature 2436 K. The value of peak wavelength obtained is 1.18955 μm. The value of radiance per m^{2} per steradian is 267,587 watt. About 5.98557e+18 photons are required for one joule of energy in 100 watt bulb.

In 100 wattage bulb, all the radiations beyond peak value of wavelength (i.e. 1.189 × 10^{−6} m) lie in the infrared region which can be seen from ^{−6} m below. About 5 percentages of the emitted radiations exists within wavelength range 0.5 μm and 0.75 μm. In total, about 8% of the emitted radiations fall in visible range.

Violet colour of light is not emitted in 100 watt bulb. Blue colour is found in extremely negligible amount in the spectrum of the visible radiations. The spectrum of other colour of light including negligible amount of blue colour, is observed in 100 watt bulb.

500 watt bulb filament gets heated at temperature 2575 K. The value of peak wavelength obtained is 1.12534 μm. The value of radiance per m^{2} per steradian is 334,091 watt. About 5.66338e+18 photons are required for one joule of energy in 500 watt bulb. The intensity of the emitted radiations is 1.04958e+06 watt in 500 watt bulb.

In 500 wattage bulb, all the radiations beyond peak value of wavelength (i.e. 1.125 × 10^{−6} m) lie in infrared region which can be known from ^{−6} m. About 13 percentage of the emitted radiations, fall between 0.75 μm and 1 μm. About “6” percentage of the emitted radiations, fall in the wavelength range, between 0.5 μm and 0.75 μm.

About 11% of the emitted radiations fall within the spectrum of the visible range. Extremely negligible amount of the wavelength of violet colour can be expected in the spectrum of the radiations of the 500 watt bulb. About 89% of the emitted radiations fall in infrared and other higher wavelength region.

When the radiation curves in Figures 10-12, of 6, 100, and 500 watt respectively are compared, the amount of infrared radiations emitted is seen greatest in 500 watt bulb and least in 6 watt bulb. The radiation curve is more flattened in 6 watt bulb and less flattened in 500 watt bulb. This is due to the lower value of peak emission wavelength in 500 watt bulb than 6 watt bulb.

The radiant emittance depends upon the intensity of the emitted radiations. Here, the 500 watt bulb has the greatest radiant emittance than the lower wattage bulb (6 watt) bulb. The spectral radiance also goes on increasing with increase in the power of the bulb. More clearly, the energy emitted by a surface into a solid angle, in a specific direction, in a unit time interval, by a unit projected area, over a unit wavelength interval is greater in higher wattage bulb (500 watt) than the lower wattage bulb (6 watt bulb). The number of photons per joule is greater in 6 wattage bulb than the 500 wattage bulb. The number of photons required for one joule for lower wattage bulb (6 watt) is greater than the higher wattage bulb (500 watt) bulb. Since, number of photons per joule (or per energy) depends directly on the wavelength of the emitted radiation, more numbers of photons are required for one joule of energy in 6 watt bulb than in 500 watt bulb. More strictly, comparatively few amounts of shorter wavelength radiations are emitted in lower wattage bulb than in higher wattage bulb. The higher value of spectral radiance in 6 watt bulb indicates that there is more distribution of longer wavelength radiations than shorter wavelength distribution. The lower value of spectral emittance in 500 watt bulb indicates that there is more distribution of shorter wavelength radiations than longer wavelength distribution.

The error between two values of the resistance for the same wattage bulb is due to the value of the resistivity. The value of resistivity of the tungsten filament varies with the temperature, but we have taken same particular value of resistivity for different wattage bulbs, which gets heated at different temperatures after passing current. For example, the difference in the value of resistance for 6 wattage bulb (obtained by two ways) is greater in comparison to the higher wattage bulb. That is because the 6 wattage bulb gets heated at a lower temperature than higher wattage bulbs. So the value of the resistivity taken in the calculation (on using R = ρ L A ) is to be lowered.

The root mean square error (RMSE) is less in power curve fitting than inverse curve fitting. The root mean square error in power curve fitting is 0.002996, and in inverse curve fitting, the root mean square error is 0.003218. The points fit more exactly in power curve. The equation for power curve is y = a x b . The value of a is 8397 and b is −0.9999. The value of x varies from 100 - 2500 nm.

In inverse curve fitting, the points are justified by the equation y = a x , where a = 8400 and RMSE is 0.003218.

Normal DistributionThe probabilities of finding the particular wavelength in the certain range are calculated below. The probabilities of finding the wavelength 400, 500, and 600 nm wavelengths of visible radiations are only discussed here and are shown in Figures 13-15 respectively. The lower and upper limits are taken 0.3 μm and 1 μm respectively. The probability depends upon the mean value, standard deviation, value of lower limit and value of upper limit. The parameters taken to calculate the probability of obtaining the mean value satisfies the result. Other results can also be verified in a similar way by varying the parameters, since all parameters (value of lower limit, value of upper limit, standard deviation, and mean value) can be taken as variables. The area under the curve represents the total probability “1”. The highlighted area under the curve represents the probability occurrence of the particular wavelength within certain range.

1) 400 nm Wavelength

Here, probability of finding 400 nm wavelengths in the range between 0.3 μm, and 1 μm is discussed. The spread value is 0.7 μm.

Lower limit (a) = 0.3 × 10^{−6} m

Upper limit (b) = 1 × 10^{−6} m

Standard deviation (σ) = 0.7 × 10^{−6}

Mean value (μ) = 400 nm

Probability = 0.35801

The blue colour area under the curve in

Total probability = probability of finding 400 nm wavelength + probability of finding other wavelength except 400 nm wavelength.

Probability of finding other wavelength except 400 nm wavelength = Total probability − probability of finding 400 nm wavelength = 1 − 0.35801

= 0.64199

= 64.199%

The probability of finding other wavelengths between 0.3 μm and 1 μm except 400 nm wavelength is 64.119%.

2) 500 nm Wavelength

Here, probability of finding 500nm wavelength in the range, between 0.3 μm, and 1 μm is discussed. The spread value is 0.7 μm.

Lower limit (a) = 0.3 × 10^{−6} m

Upper limit (b) = 1 × 10^{−6} m

Standard deviation (σ) = 0.7 × 10^{−6}

Mean value (μ) = 500 nm

Probability = 0.37141

Bluer colour areas under the curve in

Total probability = probability of finding 500 nm wavelength + probability of finding other wavelength except 500 nm wavelength.

Probability of finding other wavelength except 500 nm wavelength = Total probability − probability of finding 500 nm wavelength = 1 − 0.37141

= 0.62859

= 62.859

The probability of finding other wavelengths between 0.3 μm and 1 μm except 500 nm wavelength is 62.859%.

3) 600 nm Wavelength

Here, probability of finding 600 nm wavelengths in the range between 0.3 μm, and 1 μm is discussed. The spread value is 0.7 μm.

Lower limit (a) = 0.3 × 10^{−6} m

Upper limit (b) = 1 × 10^{−6} m

Standard deviation (σ) = 0.7 × 10^{−6}

Mean value (μ) = 600 nm

Probability = 0.37842

The probability of finding radiations of wavelength 600 nm in the limit between 0.3 μm, and 1 μm is 0.37842. About 37.842 percentages of the radiations having wavelength 600 nm exists in the range between 0.3 μm, and 1 μm.

Total probability = probability of finding 600 nm wavelength + probability of finding other wavelength except 600 nm wavelength.

Probability of finding other wavelength except 600 nm wavelength = Total probability − probability of finding 600 nm wavelength = 1 – 0.37842

= 0.62158

= 62.158%

This shows that probability of finding other wavelengths between 0.3 μm and 1 μm except 600 nm wavelength is 62.158%.

The clustered around the central peak is greatest in 600 nm wavelength and least in 400 nm wavelength. This signifies that the probability occurrence of 600 nm wavelength radiation is greatest among the 400 nm, 500 nm, and 600 nm wavelengths. White region under the curve indicates the probability occurrence of other wavelengths except those whose probability occurrence is to be determined.

The temperature measurement equation released in this work, can be used for all heated solid body (having resistance that depends upon its length and area) due to applied voltage. The peak value of wavelength obtained by using Wein’s displacement law is only an indicator which helps to investigate, “how the energy of the emitted radiations is distributed”. So, Wein’s displacement law can be used if we have to find the probability distribution of radiation energy density in filament bulb. Only about small fraction of the emitted radiations (that may vary on the power of the bulb) fall in visible region, remaining fraction may be useful in heat production. So the filament bulb is useful for both light and significant heat.

Wein’s displacement law is applicable in interpretation of electromagnetic spectrum radiated by the filament bulb. The Wein’s displacement equation that had been used is ( λ max ) = h c 4.965 k T . If we suppose the constant value in the right hand side, different than the value (4.965), keeping others parameters constant, there will be two probability of getting the value of wavelength. The two probable values are either greater, or less than the value of wavelength obtained by using the equation ( λ max ) = h c 4.965 k T . Suppose, if we lower the constant value 4.965, the value of maximum wavelength λ max , is higher. On higher value of maximum wavelength, the black body radiation curve does not give probability of obtaining wavelength in the visible region which is against the practical behavior of the filament bulb. Similarly, if we increase the constant value 4.965, the maximum wavelength is lower. On lower value of maximum wavelength, the black body distribution curve is such that there is probability of finding the smaller wavelength in the ultraviolet region which is also against the practical behavior of the filament bulb. So the equation ( λ max ) = h c 4.965 k T can be used to analyze the electromagnetic spectrum emitted from the filament bulb.

Filament bulb follows blackbody radiation curve. Only very small fractions of radiation fall in visible region. The curve is followed by radiated power density Planck law. Ultraviolet radiations are not emitted from the filament bulb of the observed watt. In “6” wattage bulb, spectrum of the emitted radiations does not contain cyan, blue and violet colour of light. In 100 watt bulb, violet colour is absent in the spectrum of the emitted visible light. In 500 watt bulb, spectrum of all colours of light is observed. Maximum intensity of red colour of light is emitted in the filament bulb. The intensity of emitted radiations goes on decreasing with decrease in wavelength of the radiations. This signifies that less number of radiations or photons of higher energy is emitted in higher wattage bulb and more number of radiations with lower energy is emitted in lower wattage bulb. For example, in Figures 10-12, the intensity of emitted red colour of light is greater than other visible colours of light (when compared among the visible colour radiations). Power of bulb depends directly on radius of the filament. Greater the radius of the filament is, greater the power of the bulb is. The calculated value of power suggests that, length and diameter of the filament which has taken in the work are quite correct.

Temperature is not the key factor for power of the bulb. The higher temperature filament may also have lower power. Radius is an important factor that determines power of the bulb. Resistance of filament has also indirect relationship with power of the bulb. Resistance for higher wattage bulb is lower, so the current flowing through higher wattage bulb is greater. The current passing through the bulb goes on increasing if power of the bulb increases.

This research was supported by Research Management Committee of Birendra Multiple Campus. My Special thanks go to Birendra Multiple Campus who provided insight and expertise that greatly assisted the research. This journey would not have been possible without the support of my family members and professors.

The author declares no conflicts of interest regarding the publication of this paper.

Rajendra, N. (2020) Study on Emitted Radiations from Filament Bulb of Different Power. Journal of Applied Mathematics and Physics, 8, 1615-1645. https://doi.org/10.4236/jamp.2020.88124