The power radiated by a black body is P and it radiates maximum energy at wavelength, \lambda 0 . If the temperature of the black body is now changed so that it radiates maximum energy at wavelength 3 \over 4\lambda _0 , the power radiated by it becomes nP. The value of n is

Properties of Matter question

2018 · Q145

Properties of Matter question

2018 · Q145

View written solutionFree

Correct answer: C

From Wien's law, $\lambda$ _maxT = constant

$\therefore$ $\lambda$ _max₁T₁ = $\lambda$ _max₂T₂

$\Rightarrow$ $\lambda$ ₀T₁ = ${{3{\lambda _0}} \over 4}$ T₂

$\Rightarrow$ ${{T_2} \over T_1} = {4 \over 3}$

According to Stefan-Boltzmann law, energy emitted
unit time by a black body is Ae $\sigma$ T⁴,

$\therefore$ P $\propto$ T⁴

So ${{{P_2}} \over {{P_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$ = ${\left( {{4 \over 3}} \right)^4}$

Here, P₂ = nP and P₁ = P

So, ${{nP} \over P} = n = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{4 \over 3}} \right)^4} = {{256} \over {81}}$

More from Properties of Matter

Sign in

Properties of Matter question

2018 · Q145

Properties of Matter question

2018 · Q145

View written solutionFree

Correct answer: C

From Wien's law, $\lambda$ _maxT = constant

$\therefore$ $\lambda$ _max₁T₁ = $\lambda$ _max₂T₂

$\Rightarrow$ $\lambda$ ₀T₁ = ${{3{\lambda _0}} \over 4}$ T₂

$\Rightarrow$ ${{T_2} \over T_1} = {4 \over 3}$

According to Stefan-Boltzmann law, energy emitted
unit time by a black body is Ae $\sigma$ T⁴,

$\therefore$ P $\propto$ T⁴

So ${{{P_2}} \over {{P_1}}} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4}$ = ${\left( {{4 \over 3}} \right)^4}$

Here, P₂ = nP and P₁ = P

So, ${{nP} \over P} = n = {\left( {{{{T_2}} \over {{T_1}}}} \right)^4} = {\left( {{4 \over 3}} \right)^4} = {{256} \over {81}}$