Radiation (General, Kirchoff's law, Black body, Prevost's Theory) Questions in English

(C) pyrometer is a type of remote-sensing thermometer used to measure the temperature of surfaces.
Specifically,a total radiation pyrometer is used to measure very high temperatures,such as those of the sun,because it does not require physical contact with the source.
It works on the principle of Stefan-Boltzmann law,which states that the total energy radiated by a black body is proportional to the fourth power of its absolute temperature $(E \propto T^4)$.
Pyrometers can typically measure temperatures ranging from $800^{\circ}C$ to $6000^{\circ}C$.
Therefore,the temperature of the sun is measured with a pyrometer.

(A) The correct answer is $A$. In the winter,the air surrounding the body is cold. When sitting in the sunshine,the solar radiation heats the air molecules near the body and the body directly absorbs radiant energy. This heat transfer from the warmer surroundings and radiation to the body makes the person feel comfortable.

(A) Heat radiation consists of electromagnetic waves,specifically infrared radiation.
In a vacuum,all electromagnetic waves travel at the speed of light,which is approximately $c = 3 \times 10^8 \ m/s$.
Therefore,the velocity of heat radiation in a vacuum is equal to the velocity of light.

(C) The rate of heat transfer depends on the mechanism of energy propagation. Conduction and convection require a material medium and involve the movement of particles or kinetic energy transfer between adjacent molecules,which is relatively slow. Radiation,however,involves the emission of electromagnetic waves,which travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$. Therefore,radiation is the fastest mode of heat transfer.

(D) The correct answer is $D$.
$A$ thermopile is a sensitive electronic device that converts thermal energy into electrical energy.
It is specifically designed for the detection of thermal radiation and the measurement of its intensity by utilizing the Seebeck effect.

(D) The inner and outer surfaces of a thermos flask are polished to minimize heat transfer by radiation. $A$ polished surface is a poor absorber and a good reflector of thermal radiation. Therefore,by polishing the surfaces,the flask reflects all incoming radiation from the outside and prevents the internal heat from radiating out,thereby maintaining the temperature of the contents.

(C) Heat transfer occurs through three main processes: conduction,convection,and radiation.
Conduction requires a material medium and involves the transfer of energy through molecular collisions.
Convection also requires a material medium (liquid or gas) and involves the actual movement of matter.
Radiation is the process of heat transfer in the form of electromagnetic waves,which do not require any material medium to propagate.
Therefore,heat can travel through a vacuum only by radiation.

(D) According to Prevost's theory of heat exchange,every body at all temperatures (above $T = 0 \ K$) continuously emits and absorbs thermal radiation.
Since the human body temperature is approximately $37^{\circ}C$ $(310 \ K)$,it emits electromagnetic radiation corresponding to its temperature.
According to Wien's displacement law,the wavelength of maximum emission for a body at $310 \ K$ falls in the infra-red region of the electromagnetic spectrum.
Therefore,the radiation emitted by the human body is in the infra-red region.

(B) Infrared radiation is electromagnetic radiation with wavelengths longer than those of visible light.
Because these radiations are associated with heat,they are commonly detected using a pyrometer,which measures the intensity of thermal radiation emitted by an object.

(A) In a vacuum,heat transfer cannot occur via conduction or convection because these processes require a material medium.
Heat transfer in a vacuum occurs exclusively through the process of radiation,which involves the emission of electromagnetic waves.
Therefore,the hot body loses thermal energy to its surroundings or the cold body through radiation,leading to a decrease in its temperature.

(C) According to Kirchhoff's law of radiation,for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium,the emissivity is equal to the absorptivity.
Therefore,good absorbers of heat are also good emitters of heat.

(A) perfectly black body is defined as an object that absorbs all the electromagnetic radiation incident upon it,regardless of frequency or angle of incidence.
By definition,the absorptive power $(a)$ of a body is the ratio of the energy absorbed to the total energy incident on it.
Since a perfectly black body absorbs all incident radiation,the energy absorbed equals the total energy incident.
Therefore,its absorptive power is $a = 1$.

(D) According to Kirchhoff's law of radiation,a body that is a good emitter of radiation at a given temperature is also a good absorber of radiation at the same temperature.
Specifically,if a substance emits certain discrete wavelengths at a high temperature,it will absorb those same specific wavelengths when it is at a lower temperature.
Therefore,if the substance emits ${\lambda _1}, {\lambda _2}, {\lambda _3}$ and ${\lambda _4}$ at high temperatures,it will absorb the same set of wavelengths,${\lambda _1}, {\lambda _2}, {\lambda _3}$ and ${\lambda _4}$,when cold.
The correct option is $D$.

(B) According to the principles of radiation,a black surface is a better absorber and a better emitter of heat radiation compared to a white surface.
$1$. Absorption: $A$ person with black skin absorbs more incident heat radiation from the surroundings,causing them to feel more heat.
$2$. Emission: $A$ person with black skin also emits more heat radiation to the surroundings,causing them to lose heat faster and feel more cold.
Therefore,a person with black skin experiences both more heat and more cold.

(A) According to Kirchhoff's law of thermal radiation,for any body in thermal equilibrium,the emissivity $e$ is equal to its absorptive power $a$ at a given wavelength and temperature.
For a perfectly black body,the absorptive power $a = 1$ and the emissivity $e = 1$.
Therefore,the relation is $e = a$.

(B) According to Kirchhoff's law of radiation,good absorbers of radiation are also good emitters (radiators) of radiation.
Rough and black surfaces are good absorbers,and therefore,they are also good radiators.
Highly polished,mirror-like surfaces are excellent reflectors of radiation,which means they are poor absorbers and consequently poor radiators.
Therefore,the statement that highly polished mirror-like surfaces are very good radiators is incorrect.

(B) The correct answer is $B$. Black surfaces are good absorbers of radiant heat from the sun. Since the black cloth absorbs more solar radiation compared to the white cloth,the thermal energy transferred to the ice underneath the black cloth is higher. Consequently,the ice covered by the black cloth will melt more than the ice covered by the white cloth.

(C) According to Kirchhoff's law of thermal radiation,for any body in thermal equilibrium,the ratio of its emissive power $(e_\lambda)$ to its absorptive power $(a_\lambda)$ at a given wavelength $\lambda$ and temperature $T$ is equal to the emissive power $(E_\lambda)$ of a perfectly black body at the same wavelength and temperature.
Mathematically,this is expressed as:
$\frac{e_\lambda}{a_\lambda} = E_\lambda$
Rearranging this equation,we get:
$e_\lambda = a_\lambda E_\lambda$
Therefore,option $C$ is the correct statement.

(B) The absorption power (or absorptivity) of a body is defined as the ratio of the amount of heat absorbed by the body to the total amount of heat incident on it.
Given:
Total heat incident on the body = $p$ calories.
Heat absorbed by the body = $q$ calories.
Therefore,Absorption power = $\frac{\text{Heat absorbed}}{\text{Total heat incident}} = \frac{q}{p}$.
Thus,the correct option is $B$.

(C) Planck's law correctly describes the spectral energy distribution of black body radiation across all wavelengths and temperatures.
While Wien's law only works for short wavelengths and Rayleigh-Jeans law works for long wavelengths,Planck's law provides a unified explanation that matches experimental observations at both low and high temperatures.

(C) On a clear night,the black seat of a bicycle loses heat rapidly by radiation because black surfaces are good radiators of heat energy.
As the seat loses heat,its temperature drops below the dew point of the surrounding air.
Consequently,the water vapour present in the air condenses on the cold surface of the seat,making it wet.

(A) According to Kirchhoff's law of radiation,good absorbers of radiation are also good emitters of radiation at a given temperature.
$1$. The rough black spot is a good absorber of radiation,and therefore,it is also a good emitter of radiation.
$2$. The polished metallic plate is a poor absorber of radiation,and therefore,it is a poor emitter of radiation.
$3$. When the plate is heated to $1400 K$ and taken into a dark room,the black spot emits a significant amount of thermal radiation,making it appear bright.
$4$. The polished surface emits very little radiation,making it appear dark in comparison to the spot.
Therefore,the spot will shine more than the plate.

(B) According to Kirchhoff's law of radiation,at a certain temperature and for a given wavelength,the ratio of the emissive power $(E_{\lambda})$ of a body to the emissive power $(E_{b\lambda})$ of a black body at the same temperature is defined as the emissivity (or spectral emissivity) of the body.
Mathematically,$\epsilon_{\lambda} = \frac{E_{\lambda}}{E_{b\lambda}}$.
Therefore,the correct option is $B$.

(A) black body is a better absorber of heat radiation compared to a non-black body.
Since the bulb of thermometer $A$ is painted black,it absorbs heat radiation from the sunlight at a faster rate than thermometer $B$.
Therefore,the temperature of $A$ rises faster than $B$.
However,once both thermometers reach thermal equilibrium with the surrounding environment,they will both eventually attain the same final atmospheric temperature.

(C) According to Kirchhoff's law,good absorbers of radiation are also good emitters of radiation at a given temperature. Since a black spot is a good absorber of light and heat,it will also be a good emitter. Therefore,when the body is heated and placed in a dark room,the black spot will emit more radiation than the surrounding surface,making it appear to glow more brightly.

(A) According to Kirchhoff's law of radiation,good absorbers of a particular wavelength are also good emitters of that same wavelength.
Red glass appears red because it transmits red light and absorbs green light (as red and green are complementary colors).
When the red glass is heated to a high temperature and placed in a dark room,it emits the radiation it is capable of absorbing most strongly.
Since red glass is a strong absorber of green light,it will emit green light strongly when heated. Therefore,it will appear green in a dark room.

(D) According to $Kirchoff's$ law of radiation,the emissivity of a body is equal to its absorptivity at a given temperature.
Black and rough surfaces are excellent absorbers of heat radiation.
Consequently,they are also the most efficient emitters of heat radiation.
Therefore,a hot body will radiate heat most rapidly if its surface is black and rough.

(D) perfectly black-body is defined as an ideal body that absorbs all incident radiation of any wavelength falling on it. According to Kirchhoff's law of thermal radiation,a good absorber is also a good emitter. Therefore,a perfectly black-body is also a perfect emitter,meaning it emits radiation of all possible wavelengths at a given temperature.

(C) An ideal black body is defined as a body that absorbs all incident radiation of any wavelength.
When light is incident on a pin hole,it enters the box and undergoes successive reflections at the inner walls.
At each reflection,a portion of the energy is absorbed by the walls.
Consequently,once the ray enters the box,it is almost impossible for it to escape,causing the pin hole to act as a perfect black body.

(A) Initially,an ideal black body absorbs all the radiant energy incident on it,making it appear as the darkest object in the furnace.
According to Kirchhoff's law of radiation,a good absorber is also a good emitter. Therefore,as the black body heats up to the temperature of the furnace,it radiates the maximum amount of energy compared to other bodies at the same temperature.
Once the black body reaches thermal equilibrium with the furnace,it emits the most intense radiation,making it appear the brightest.

(D) According to Kirchhoff's law,a good absorber is a good emitter,so option $(A)$ is incorrect.
Every body absorbs and emits radiation at all temperatures above $0 \, K$ (absolute zero). At $0 \, K$,thermal motion ceases,so option $(B)$ is incorrect.
The energy of radiation emitted from a black body varies with wavelength,as described by the black body radiation spectrum,so option $(C)$ is incorrect.
Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$. The relation is given by $E_{\lambda} d_{\lambda} = \frac{8\pi hc}{\lambda^5} \frac{1}{[e^{hc/\lambda kT} - 1]} d_{\lambda}$. Therefore,option $(D)$ is correct.

(C) According to Kirchhoff's law of radiation,good absorbers are good emitters at a given temperature.
Blue glass absorbs all wavelengths except blue at room temperature. When heated to a high temperature,it becomes a good emitter of the wavelengths it previously absorbed (which are the longer wavelengths,i.e.,red end of the spectrum).
When taken into a dimly lit room,the heated blue glass emits these longer wavelengths,making it appear to glow with a bright red color.
The red glass at room temperature does not emit significant visible radiation.
Therefore,the blue piece shines like a brighter red compared to the red piece.

(B) Kirchhoff's law states that the ratio of emissive power to the absorptive power corresponding to a particular wavelength and at a given temperature is constant for all bodies:
$e_{\lambda} / a_{\lambda} = \text{constant} = E_{\lambda}$
where $E_{\lambda}$ is the emissive power of a perfect black body corresponding to the same wavelength and temperature. It is clear from the above relation that if $a_{\lambda}$ (absorptive power) is large for a body, $e_{\lambda}$ (emissive power) will also be large to keep the ratio constant, i.e., good absorbers are good emitters.

(B) According to Prevost's theory,every body radiates heat at all temperatures (except $0 \ K$) and also absorbs heat from its surroundings.
Since $T_A > T$,the rate of emission of radiation by object $A$ is greater than the rate of absorption of radiation from the surroundings. Thus,$A$ loses more radiation than it absorbs.
Since $T_B < T$,the rate of absorption of radiation by object $B$ from the surroundings is greater than the rate of emission of radiation by it. Thus,$B$ absorbs more radiation than it emits.
Eventually,both objects reach thermal equilibrium with the surroundings at temperature $T$.

(B) According to Prevost's theory of heat exchange,every body at a temperature above $0 \ K$ radiates thermal energy to its surroundings and simultaneously absorbs thermal energy from its surroundings.
When the temperature of the body $(T_b)$ is equal to the temperature of the surroundings $(T_s)$,the rate of emission of heat energy is exactly equal to the rate of absorption of heat energy.
Therefore,the net exchange of heat is zero,and the body remains in thermal equilibrium with its surroundings.
Thus,the correct statement is that it radiates the same quantity of heat as it absorbs.

(A) The rate of cooling is given by the formula $\frac{d\theta}{dt} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.
Since all objects have the same mass $m$,material (same specific heat $c$),and are at the same initial temperature $T$,the rate of cooling is directly proportional to the surface area $A$ of the object $(\frac{d\theta}{dt} \propto A)$.
For a given mass and density,the surface area is largest for a thin circular plate and smallest for a sphere.
Therefore,the plate has the largest surface area,which leads to the fastest rate of cooling.
Thus,the plate will cool first.

(A) According to Stefan-Boltzmann law,the rate of heat loss is given by $\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$.
Since $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,we have $-ms \frac{dT}{dt} = e \sigma A (T^4 - T_0^4)$.
Thus,the rate of cooling $\left( -\frac{dT}{dt} \right) = \frac{e \sigma A}{ms} (T^4 - T_0^4)$.
For bodies of the same surface area and mass,the rate of cooling is directly proportional to the emissivity $(e)$,i.e.,$\left( -\frac{dT}{dt} \right) \propto e$.
From the graph,the slope of the curve for body $x$ is steeper than that for body $y$ at any given temperature,which means $\left( -\frac{dT}{dt} \right)_x > \left( -\frac{dT}{dt} \right)_y$.
Therefore,$e_x > e_y$.
According to Kirchhoff's law of radiation,for any body,the emissivity $(e)$ is equal to its absorptive power $(a)$ at a given wavelength and temperature,i.e.,$e = a$.
Thus,$e_x > e_y$ implies $a_x > a_y$.

(C) According to Wien's displacement law,the wavelength corresponding to maximum energy emission is inversely proportional to the absolute temperature: $\lambda_m \propto \frac{1}{T}$.
Since frequency $\nu$ is related to wavelength $\lambda$ by $\nu = \frac{c}{\lambda}$,we have $\nu_m \propto T$.
This implies that as the temperature $T$ of the black body increases,the frequency $\nu_m$ corresponding to the maximum energy emission also increases.
Looking at the provided graphs,as the temperature increases from $1500 \ K$ to $2500 \ K$ to $3500 \ K$,the peak of the curve should shift towards higher frequencies (to the right).
Graph $(c)$ correctly depicts this behavior,where the peak frequency increases with increasing temperature.

(A) According to Wien's displacement law,the wavelength corresponding to maximum intensity $\lambda_m$ is inversely proportional to the absolute temperature $T$,i.e.,$\lambda_m \propto \frac{1}{T}$.
Given $T_2 > T_1$,it follows that $\lambda_{m_2} < \lambda_{m_1}$.
This means the peak of the intensity-wavelength curve for the higher temperature $T_2$ shifts towards a shorter wavelength (left side) compared to the curve for $T_1$.
Additionally,according to the Stefan-Boltzmann law,the total energy emitted per unit area per unit time is proportional to $T^4$,so the intensity $I$ for $T_2$ will be higher than for $T_1$ at all wavelengths.
Therefore,the correct plot is the one where the curve for $T_2$ is higher and its peak is shifted to the left relative to the curve for $T_1$.

(B) thermopile is a device that converts thermal energy into electrical energy. It consists of several thermocouples connected in series. It is primarily used for the detection and measurement of radiant heat energy (infrared radiation) by producing a voltage proportional to the temperature difference.

(B) Kirchhoff's Law states that good emitters are good absorbers at a given temperature.
In the solar spectrum,the dark lines known as Fraunhofer lines are formed because the cooler gases in the Sun's outer atmosphere (chromosphere) absorb specific wavelengths of light emitted by the hotter inner layers (photosphere).
During a solar eclipse,when the bright photosphere is obscured by the Moon,the chromosphere becomes visible. At this moment,the emission spectrum of the chromosphere is observed,showing bright lines at the exact same wavelengths as the dark Fraunhofer lines.
This phenomenon is a direct consequence of Kirchhoff's law,which explains why the same material that absorbs light at a specific wavelength will also emit light at that same wavelength.

(C) The intensity of radiation on a surface is given by the formula $I_{\theta} = I_0 \cos \theta$,where $I_0$ is the intensity when the surface is normal to the rays and $\theta$ is the angle between the normal to the surface and the direction of the rays.
Given $\theta = 60^{\circ}$,we have:
$I_{\theta} = I_0 \cos 60^{\circ}$
Since $\cos 60^{\circ} = \frac{1}{2}$,we get:
$I_{\theta} = I_0 \times \frac{1}{2} = \frac{I_0}{2}$.

(C) The correct option is $C$.
Heat radiations,also known as infrared radiations,are a form of electromagnetic waves.
According to the properties of electromagnetic waves,they travel through a vacuum at the speed of light,which is approximately $3 \times 10^8 \ m/s$.
Since heat radiations are electromagnetic in nature,they propagate with the same speed as light waves.

(C) The surface temperature of the Sun is approximately $6000 \, K$.
This value is widely accepted in astrophysics as the effective temperature of the solar photosphere.

(D) For any surface,the sum of absorptance $(a)$,reflectance $(r)$,and transmittance $(t)$ is equal to $1$,i.e.,$a + r + t = 1$.
Given,$a = 0.4$ and $r = 0.6$.
Substituting these values into the equation: $0.4 + 0.6 + t = 1$.
$1.0 + t = 1$.
Therefore,$t = 0$.
Since the transmittance $(t)$ is $0$,no radiation can pass through the body.
Thus,the body is perfectly opaque.

(B) According to $Prevost's$ theory of heat exchange,every body at a temperature above $0 \ K$ emits and absorbs thermal radiation simultaneously.
When a body is in thermal equilibrium with its surroundings,its temperature remains constant.
This implies that the rate of heat energy absorbed by the body must be exactly equal to the rate of heat energy emitted by the body.
Therefore,the net heat exchange is zero,and the body maintains a constant temperature equal to that of the surroundings.

(A) According to the Stefan-Boltzmann law,the total energy $E$ radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^{4}$.
According to Wien's displacement law,the wavelength $\lambda_{max}$ corresponding to maximum energy emission is inversely proportional to temperature: $\lambda_{max} \propto 1/T$.
The spectral energy density $E_{\lambda}$ is given by Planck's law. The maximum spectral energy density $E_{max}$ at a given temperature $T$ is proportional to the fifth power of the absolute temperature: $E_{max} \propto T^{5}$.
Therefore,the correct relationship is $E \propto T^{4}$ and $E_{max} \propto T^{5}$.

(C) According to Kirchhoff's law of radiation,the ratio of emissive power $(e)$ to the absorptive power $(a)$ for a body at a given temperature and wavelength is constant and equal to the emissive power of a perfectly black body at that temperature.
Mathematically,$\frac{e}{a} = E_{\lambda}$,where $E_{\lambda}$ is the emissive power of a black body.
This implies that $e \propto a$.
$A$ black spot on a body has a higher absorptive power $(a \approx 1)$ compared to the rest of the surface.
Since it is a better absorber,it is also a better emitter of radiation at high temperatures.
Therefore,when the body is heated and placed in a dark room,the black spot emits more radiation and appears to glow more brightly.
Thus,the correct option is $C$.

(D) Prevost's theory of heat exchange states that all bodies at all temperatures (above $0 \ K$) emit and absorb thermal radiation simultaneously.
When we enter an air-conditioned room,our body temperature is higher than the room temperature.
According to Prevost's theory,the rate of emission of heat from our body is greater than the rate of absorption of heat from the surroundings.
This net loss of heat from our body results in the sensation of feeling cold.

(C) line spectrum is produced by excited atoms and consists of discrete wavelengths.
$A$ band spectrum consists of groups of closely spaced lines that appear as bands,typically produced by molecules.
$A$ continuous spectrum is one that contains all wavelengths within a certain range without any gaps or interruptions.
Black body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment. According to Planck's law,a black body emits radiation across a continuous range of wavelengths,which depends solely on its temperature. Therefore,the spectrum of black body radiation is a continuous spectrum.