Question

The dimensions of Stefan-Boltzmann constant $\sigma $ can be written in terms of planck's constant $h$ , Boltzmann constant ${K_B}$ and the speed of light $c$ as $\sigma = {h^\alpha }K_B^\beta {c^\gamma }$. Here,
A. $\alpha = 3,\beta = 4,and\text{ } \gamma = - 3$
B. $\alpha = 3,\beta = - 4,and\text{ } \gamma = 2$
C. $\alpha = - 3,\beta = 4,and\text{ } \gamma = - 2$
D. $\alpha = 2,\beta = - 3,and\text{ } \gamma = - 1$

Answer 1

Hint: In this question, we start by writing the given equation in the dimensional form $ \Rightarrow \left[ \sigma \right] = {\left[ h \right]^\alpha }{\left[ {{K_B}} \right]^\beta }{\left[ c \right]^\gamma }$ then we write the dimension of each of the constant that is the dimension of Stefan-Boltzmann constant $\sigma $ is \[\left[ {M{T^{ - 3}}{K^{ - 4}}} \right]\], planck's constant $h$ is\[\left[ {M{L^2}{T^{ - 3}}} \right]\], Boltzmann constant ${K_B}$ is \[\left[ {M{L^2}{T^{ - 2}}{K^{ - 1}}} \right]\], and speed of light $c$ is \[\left[ {L{T^{ - 1}}} \right]\]. After we substitute these in the dimensional equation given above and compare the powers of T, M, K, and L . Then we get the values of power as
$\alpha = - 3,\beta = 4,and\gamma = - 2$.

Complete Step-by-Step solution:
We start by equating the dimensions of left hand side and the dimensions of the right hand side that is
$\sigma = {h^\alpha }K_B^\beta {c^\gamma }$
$ \Rightarrow \left[ \sigma \right] = {\left[ h \right]^\alpha }{\left[ {{K_B}} \right]^\beta }{\left[ c \right]^\gamma }$------------------------------- (1)
Now we need to write the dimensions of all the constant in equation (1), that is

1. The dimension of Stefan-Boltzmann constant $\sigma $ is given as \[\left[ {M{T^{ - 3}}{K^{ - 4}}} \right]\] by using Stefan’s law that is $\dfrac{Q}{{At}} = \sigma {T^4}$

2. The dimension of planck's constant $h$ is given as \[\left[ {M{L^2}{T^{ - 3}}} \right]\] by using $E = h\gamma $

3. The dimension of Boltzmann constant ${K_B}$ is given as \[\left[ {M{L^2}{T^{ - 2}}{K^{ - 1}}} \right]\]by using $E = \dfrac{3}{2}{K_B}T$

4. The dimension of the speed of light $c$is given as \[\left[ {L{T^{ - 1}}} \right]\]
Now substituting the above dimension in equation number (1) we get
$ \Rightarrow \left[ {M{T^{ - 3}}{K^{ - 4}}} \right] = {\left[ {M{L^2}{T^{ - 3}}} \right]^\alpha }{\left[ {M{L^2}{T^{ - 2}}{K^{ - 1}}} \right]^\beta }{\left[ {L{T^{ - 1}}} \right]^\gamma }$------------ (2)
Now first comparing the power of $K$ from RHS and LHS of equation (2), we get
\[ \Rightarrow - \beta = - 4\]
\[ \Rightarrow \beta = 4\]--------------------------------------------- (3)
Now comparing the power of $M$ from RHS and LHS of equation (2), we get
$ \Rightarrow 1 = \alpha + \beta $------------------------------------------ (4)
Now substituting equation (3) in equation (4) we get
$ \Rightarrow 1 = \alpha + 4$
$ \Rightarrow \alpha = 1 - 4 = - 3$--------------------------------------- (5)
Now comparing the power of $L$ from RHS and LHS of equation (2), we get
$ \Rightarrow 2 = 2\alpha + 2\beta + \gamma $------------------------------------- (6)
Substituting the value of $\alpha {\text{ and }}\beta $in equation (6) we get
$ \Rightarrow 0 = 2\left( { - 3} \right) + 2\left( 4 \right) + \gamma $
$ \Rightarrow 0 = - 6 + 8 + \gamma $
$ \Rightarrow \gamma = - 2$
So we get $\alpha = - 3,\beta = 4$, and $\gamma = - 2$, hence option C is correct.

Note: For these types of questions we need to remember the dimensions of some famous constants like and if that is not possible remember the dimensional formulas for energy, force, power, and then we can derive the dimensions of constant. After that equate the power of T, L, M, and K from the dimensional formula of LHS and RHS. Then we will get our answer.