binomial formula

Q. 6. Explain Stefan's law of thermal radiation. (1996,2000,01,04)Ans. Stefan's Law : This law states that the radiant energy '£' emitted from the unit area of surface of a perfect black-body in unit time is proportional to the 4th powerof the absolute temperature of the body. -i.e. Â£ oc T4 or £ = oT4 _Where c is a constant which is called Stefan's constant and its value is 5 • 67 x 10~8 j/m2-s-K4.Suppose that a black-body having absolute temperature T\ is surrounded by a black enclosure at absolute temperature Tz- The body will emit a7|4 joule energy per second per unit area of its surface. At the same time it will absorb <rlf joule energy per second per unit area from the surrounding atmosphere. Thus net energy emitted by unit area of the body in unit time will be£1-£2=o(rl4-T24)The rate of emission of radiant energy from the area A of a body with emissive power e, at the absolute temperature T is given byE = oT4eAQ. 7. Establish Newton's law of cooling from Stefan's law. (1996,2000,01,04)Ans. Newton's Law of Cooling : This law states that if the temperature difference between a body and its surrounding is small then rate of cooling of the body (or rate of loss of heat by the body) is proportional to the temperature difference between the body and surroundings.i.e. rate of loss of heat « temperature differenceDerivation of Newton's Law From Stefan's Law: Suppose that a body is placed in air for cooling. Let the absolute temperature of the body and that of surroundings be Ti and T. Let e be the emissive power of the body. Then according to Stefan's law the net rate of loss of heat by the body is given by.A£ = ae(Ti4-T4)Let Ti-T + t wheret = 7} -T > 0is smallthen AE = ce[(T + t)i-T4]= aeT4 jl + 4—.......j -1 (Using Binomial Expansion)t < T, hence ^ is very small, hence higher degrees of ^ are neglected.= 4oeT3fv Atmosphere is very big as compared to the body, hence its temperature remains almost constant.Hence, rate of heat loss oc t (temperature difference)Q. 8. Draw spectral distribution curves of black-body radiation and write the effect of rise in temperature on it. (1996,2001)Or Draw the curves between energy and wavelength at different temperatures of a black-body radiations and write the conclusions drawn from these curves. (2003)Ans. Spectral Distribution of Black-Body Radiation : A perfect black-body, when heated upto high temperature, emits radiation of all possible wavelengths, that is why it is called full radiator. In 1899, two scientists Lummer and Pringsheim studied the spectral energy distribution in black-body radiation at differenl temperatures. They heated the black-body at differenl temperatures and at each temperature they plotted the radiant energy, emitted at different wavelengths againsl wavelength. The general shape of these curves are shown in the fig. 13 -2.Following conclusions were drawn from these graphs:(i) At a given temperature, the radiant energy 'E\ emitted by the black-body at wavelength 'A.' first increases with the wavelength and attains a maxima at a certain wavelength and then goes on decreasing.(ii) At a given temperature, maximum energy is emitted at a certain wavelength which is denoted by Xm.(iii) As the temperature is increased, the energy 'Ex emitted at wavelength X is also increased for each wavelength.(iv) The product of absolute temperature T of black-body and wavelength X,„ at which maximum energy is emitted, is a constant.i.e. XmT = b (constant)This law is called Wein's displacement law and the constant 'b' is called Wein's constant.From this law we can observe that, at low temperatures the maximum energy is emitted at larger wavelengths, but as the temperature is increased more energy is emitted in shorter wavelengths and maximum energy peak is shifted towards shorter wavelengths (see graphs).(iv) The area under these curves increases with increasing temperature directly as the 4th power of the absolute temperature of the body. That means the total radiant energy (measured by the area under the curve) emitted by the body is directly proportional to 4th power of the absolute temperature.i.e. ExT4Thus Stefan's law is also verified by these curves.Q. 9. Explain Planck's hypothesis of radiation. Discuss its importance in modern physics. (1997,98,99)Or State Planck's hypothesis. (2004)Ans. Planck's Hypothesis: According to this hypothesis, emission or absorption of radiant energy is not continuous process but is takes place in the form of small packets of energy. Each such packet is called a photon or quanta. The energy associated with a photon of radiation having frequency v is hv. Here 'h' is a constant called Planck's constant. Thus we can conclude that energies emitted by a body can be hv, 2hv,3hv....(integer multiple of hv) but not in between.Classical mechanics and thermo-dynamics could not explain the spectral energy distribution in black-body radiation. Planck, used his hypothesis and derive a formula to explain the spectral energy distribution in black-body radiation and found it in close agreement with the results of Lummer and Pringsheim experiments. Further Einstein successfully explained photo-electric effect on the basis of Planck's hypothesis. In this way Planck's hypothesis gained recogni tion.Q. 10. Find the expression for kinetic mass and momentum of photon using Planck's hypothesis. (1999,2004)Ans. Momentum of Photon: According to Planck's hypothesis, the rest mass mo of photon is zero and each photon travels with the speed of light c.If v is the frequency of photon, then its energy will be E = hv. Further if m is the kinetic mass of photon then by mass energy relation, we haveE = mc2.2 ' hv h f v 1)E = hv = mc => m = = — '•' ~ = -c2 cX V c X):. Kinetic mass of photon m = ~= —v c2cXIf p is the linear momentum of photon, thenp = kinetic mass x velocityor p=mc or p = (v velocity = c)h hvor p = — = —c/v c (â– -• X = c/v)Momentum p = — = —y X c