
In the M K S system, Stefan’s constant is denoted by \[\sigma \] . In C G S system multiplying factor of \[\sigma \] will be
A. $1$
B. ${10^3}$
C. ${10^5}$
D. ${10^2}$
Answer
510.9k+ views
Hint: The Stefan Boltzmann constant also called Stefan’s constant is the total intensity radiated over all wavelengths increases as the temperature increases of a black body which is proportional to the fourth power of the thermodynamic temperature. In this problem the M K S unit of the Stephan’s constant should be converted to a C G S unit.
Complete step by step answer:
Stephan’s constant is given by, $\dfrac{{Energy}}{{Area \times time \times temperature}}$ . …………. $\left( 1 \right)$
In M K S system, the unit of Stephan constant \[\left( \sigma \right)\] is $\dfrac{J}{{{m^2} \times \sec \times {K^4}}}$ ……….. $\left( 2 \right)$
On converting S I unit to C G S unit is
We know that,
1 joule = ${10^7}$ erg.
$1\,{m^2} = {10^4}c{m^2}$ .
On substituting in equation $\left( 2 \right)$
We get, the C G S unit of Stephan constant as,
$\dfrac{{{{10}^7}erg}}{{{{10}^4}c{m^2} \times \sec \times {K^4}}}$
On simplifying the above equation, we get
$\dfrac{{{{10}^{7 - 4}}erg}}{{c{m^2} \times \sec \times {K^4}}}$
$\Rightarrow \dfrac{{{{10}^3}erg}}{{c{m^2} \times \sec \times {K^4}}}$
Therefore, the multiplying factor of Stephen factor \[\sigma \] in C G S is ${10^3}$.
Hence, the correct option is B.
Additional information: Dimensional formula for Stefan Boltzmann constant is
On using equation $\left( 1 \right)$
Stefan constant = \[\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]\left[ T \right]\left[ {{K^4}} \right]}}\]
Therefore, Stefan's constant dimensional formula = $\left[ {M{L^0}{T^{ - 3}}{K^{ - 4}}} \right]$.
Some of the applications of Stefan Boltzmann constant are:
-Stefan Boltzmann constant is used to measure the amount of heat that is radiated from the black body.
-Stefan Boltzmann constant is used to convert temperature $\left( K \right)$ to units for intensity $\left( {W{m^{ - 2}}} \right)$ that is basically power per unit area.
Note: It should be noted that the unit for time is same in both M K S and C G S unit that is seconds and also unit for temperature is same in both M K S and C G S unit that is Kelvin and the value of Boltzmann constant is $8.617 \times {10^{ - 5}}eV{K^{ - 1}}$.
Complete step by step answer:
Stephan’s constant is given by, $\dfrac{{Energy}}{{Area \times time \times temperature}}$ . …………. $\left( 1 \right)$
In M K S system, the unit of Stephan constant \[\left( \sigma \right)\] is $\dfrac{J}{{{m^2} \times \sec \times {K^4}}}$ ……….. $\left( 2 \right)$
On converting S I unit to C G S unit is
We know that,
1 joule = ${10^7}$ erg.
$1\,{m^2} = {10^4}c{m^2}$ .
On substituting in equation $\left( 2 \right)$
We get, the C G S unit of Stephan constant as,
$\dfrac{{{{10}^7}erg}}{{{{10}^4}c{m^2} \times \sec \times {K^4}}}$
On simplifying the above equation, we get
$\dfrac{{{{10}^{7 - 4}}erg}}{{c{m^2} \times \sec \times {K^4}}}$
$\Rightarrow \dfrac{{{{10}^3}erg}}{{c{m^2} \times \sec \times {K^4}}}$
Therefore, the multiplying factor of Stephen factor \[\sigma \] in C G S is ${10^3}$.
Hence, the correct option is B.
Additional information: Dimensional formula for Stefan Boltzmann constant is
On using equation $\left( 1 \right)$
Stefan constant = \[\dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]\left[ T \right]\left[ {{K^4}} \right]}}\]
Therefore, Stefan's constant dimensional formula = $\left[ {M{L^0}{T^{ - 3}}{K^{ - 4}}} \right]$.
Some of the applications of Stefan Boltzmann constant are:
-Stefan Boltzmann constant is used to measure the amount of heat that is radiated from the black body.
-Stefan Boltzmann constant is used to convert temperature $\left( K \right)$ to units for intensity $\left( {W{m^{ - 2}}} \right)$ that is basically power per unit area.
Note: It should be noted that the unit for time is same in both M K S and C G S unit that is seconds and also unit for temperature is same in both M K S and C G S unit that is Kelvin and the value of Boltzmann constant is $8.617 \times {10^{ - 5}}eV{K^{ - 1}}$.
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