Laws of Radiation MCQ Quiz in मराठी - Objective Question with Answer for Laws of Radiation - मोफत PDF डाउनलोड करा

Concept:

The heat transfer by radiation is given by,

The Stefan Boltzman law:

Q = σ .A.ϵ .T⁴

Where, Q is heat transfer in kJ, A is heat transfer area in m²

ϵ is the emissivity of the surface and T is the temperature of the body in K.

Calculation:

Given:

Diameter of large ball d₁ = 2d and diameter of small ball d₂ = d and T₁ = T₂ for same material ϵ₁ = ϵ₂

\(A_1 =\frac{\pi }{4}\times d_1^2\) and \(A_2 =\frac{\pi }{4}\times d_2^2\)

The heat loss or cooling \(\frac{Q_1}{Q_2}= (\frac{d_1}{d_2})^2=2^2 =4\)

thus Q₁:Q₂ = 4 : 1.

Concept:

Wein's law: states that the black body radiation curve for different temperatures peak at a wavelength is inversely proportional to temperature

λ_maxT = C

Calculation:

Given:

λmT = C, C = 2898 μm.K,

From the graph, the total hemispherical emissivity will be highest at λ = 6000 Å 

λ_max = 6000 × 10-10 m / 10-6 = 0.6 μm

λmT = C ⇒ 0.6 × T = 2898

T = 2898/0.6 = 4830 K

Concept:

From the Reciprocity theorem,

\(\mathop A\nolimits_1 \mathop F\nolimits_{1 - 2} = \mathop A\nolimits_2 \mathop F\nolimits_{2 - 1} \)

Calculation:

A₁ = 6 m², A₂ = 4 m², F_1-2 = 0.1

From the Reciprocity theorem,

\(\mathop A\nolimits_1 \mathop F\nolimits_{1 - 2} = \mathop A\nolimits_2 \mathop F\nolimits_{2 - 1} \)

,\(\begin{array}{l} 6 \times 0.1 = 4 \times \mathop F\nolimits_{2 - 1} \\ \mathop F\nolimits_{2 - 1} = \frac{0.6}{4} = 0.15 \end{array}\) 

Concept:

Wien's displacement law: 

The wavelength of thermal radiation emitted by a black body is inversely proportional to the body's absolute temperature.

\(λ _{m}T= Constant\)​

where λ_m = wavelength of radiation and  T = Absolute temperature.

Calculation:

Given:

Applying Wien's law,

\(λ _{m1}T_{1}= λ _{m2}T_{2}\)

λ_m × 2000= λ × 3000

by solving,

λ = \(\frac 23\) λ_m

Hence option 2 is the correct answer.

Explanation:-

Wein's law - 

• Wein’s Law (also called Wein’s Displacement Law) is defined as so:- For a blackbody (or star), the wavelength of maximum emission of anybody is inversely proportional to its absolute temperature (measured in Kelvin).
• As a result, as the temperature rises, the maximum (peak) of the radiant energy shifts toward the shorter wavelength (higher frequency and energy) end of the spectrum (bluer).

⇒ \(λ _{max}T = 2900\ \mu m.K\;\)

The wavelength λ_max  is where the intensity is a maximum;

T is the star’s average surface temperature measured in Kelvin, and 2900 (micrometers x Kelvin) is known as Wien’s constant.

• At any wavelength, a hotter object radiates more energy (is more luminous or brighter) at all wavelengths than a cooler one.
• Maximum spectral emissive power decreases with a decrease in temperature.
• Wein’s Law tells us where (meaning at what wavelength) the star's brightness is at a maximum. We can easily see the picture below – the red dot under the word “visible” is the peak for the 6000 K object. In other words, Wien's law tells us what color the object is brightest at.

• The curves above are Planck curves for objects at different temperatures.
• Planck's law tells us how intense the thermal emission is at each wavelength.
• The peak of this Planck curve (the red dot) is what Wein’s Law tells us.
• Maximum spectral emissive power is displaced to shorter wavelengths with an increase in temperature.
• We can analyze the radiation coming from stars and obtain the energy distribution over the wavelengths of radiation emitted by them. The wavelength corresponding to the maximum energy of radiation is related to the temperature of the radiating body by the relation.
• This is Wien's displacement law and can be used to determine the temperature of stars.
• Hence the correct answer is option 1.

Explanation:

Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T.

Planck’s law for the energy Eλ radiated per unit volume by a cavity of a blackbody in the wavelength interval λ to λ + Δλ can be written in terms of Planck’s constant (h), the speed of light (c = λ × v), the Boltzmann constant (k), and the absolute temperature (T):

Energy per unit volume per unit wavelength:

\({E_\lambda } = \frac{{8\pi hc}}{{{\lambda ^5}}} \times \frac{1}{{{e^{\frac{{hc}}{{kT\lambda }} - 1}}}}\)

Energy per unit volume per unit frequency:

\({E_\nu } = \frac{{8\pi h}}{{{c^3}}} \times \frac{{{\nu ^3}}}{{{e^{\frac{{hv}}{{kT}} - 1}}}}\)

So Planck’s distribution function:

\(E\left( {\omega ,T} \right) = \frac{1}{{{e^{\frac{{h\omega }}{\tau }}} - 1}}\)

Using planck’s law, when we plot Ebλ with λ, we get the curve as shown below.

As temperature increases, the peak of the curve shift towards a lower wavelength.

Concept:

Planck’s constant: It is a physical constant that is the quantum of electromagnetic action. It relates the energy carried by a photon to its frequency by, E = hν.

\(\therefore h=\frac{E}{\nu }\)

Hence, the unit of Planck's constant is    

\(h=\frac{Joule}{1/sec}=Joule\cdot sec\)

The unit of Planck's constant is Joule⋅sec and it value is 6.6 × 10^-34 Js.

Where, 

E = energy, 

ν = frequency

h = Planck’s constant.

Explanation:

From the above explanation, we can see that, Planck's constant is a fundament constant used to define quantum energy.

And its value is approximately 6.6 × 10-34 Js.

Hence option 3 is correct among all

Concept:

According to Wein’s displacement law, the product of wavelength (λmax) corresponding to the maximum monochromatic emissive power and the absolute temperature of a black body (T) is constant.

λmax × T = constant i.e. 2900 μm-K

i.e. λ1T1 = λ2T2

Calculation:

Given T = 2000 K, λ_max = ?

Explanation:

Kirchhoff’s law:

• Let E be the total emissive power of the body and α be the absorptivity of the body.
• Emissivity (ϵ) is the ratio of the emissive power of a non-black body to the black body.
• ϵ = \(\frac{E}{E_b}\)
• Kirchhoff’s law also holds for monochromatic radiation, for which
• \(\frac{{{E_{\lambda 1}}}}{{{\alpha _{\lambda 1}}}} = \frac{{{E_{\lambda 2}}}}{{{\alpha _{\lambda 2}}}} = \frac{{{E_{b\lambda }}}}{{{\alpha _{b\lambda }}}} = \frac{{{E_{b\lambda }}}}{1}\)
• ∵ Absorptivity of a black body is one.
• \(\therefore \frac{{{E_\lambda }}}{{{\alpha _\lambda }}} = {E_{b\lambda }} \Rightarrow {\alpha _\lambda } = \frac{{{E_\lambda }}}{{{E_{b\lambda }}}} = {\epsilon_\lambda }\)
• Therefore, the monochromatic emissivity of a black body is equal to the monochromatic absorptivity at the same wavelength.
• Eλ = αλEbλ
• According to Kirchhoff’s law, the ratio of total emissive power to absorptivity is constant for all bodies which are in thermal equilibrium with the surroundings. Therefore, it means that the emissivity of a body is equal to its absorptivity.

Stefan-Boltzmann Law

• The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:
• E ∝ T4
• E = σT4
• σ = The Stefan – Boltzmann constant = 5.67 × 10^-8 Wm^-2K^-4

Wien’s displacement law

• For a black body emissive spectrum, the wavelength λmax giving the maximum emissive power at a temperature is inversely proportional to the absolute temperature.
• λ_max = \(\frac{c}{T}\)
• λ_maxT= c
• Wavelength = \(\frac{Speed ~of ~Light}{Frequency}\) ⇒ λ = \(\frac{c}{V}\)

Concept:

Emissive power is given as,  \(E = \sigma A \mathop T\nolimits^4 \)

Calculation:

Given:

T₁ = 273 + 27 = 300 K, T₂ = 927 + 273 = 1200 K

\(\frac{{\mathop E\nolimits_2 }}{{\mathop E\nolimits_1 }} = \mathop {\left( {\frac{{\mathop T\nolimits_2 }}{{\mathop T\nolimits_1 }}} \right)}\nolimits^4 = \mathop {\left( {\frac{{1200}}{{300}}} \right)}\nolimits^4 = 256\)