Dimensions of Stefan-Boltzmann's constant sigma can be written in terms of Planck's constant

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1.Units, Dimensions and Measurement

hard

The dimensions of Stefan-Boltzmann's constant $\sigma$ can be written in terms of Planck's constant $h$, Boltzmann's constant $k_B$ and the speed of light $c$ as $\sigma=h^\alpha k_B^\beta c^\gamma$. Here,

$\alpha=3, \beta=4$ and $\gamma=-3$

$\alpha=3, \beta=-4$ and $\gamma=2$

$\alpha=-3, \beta=4$ and $\gamma=-2$

$\alpha=2, \beta=-3$ and $\gamma=-1$

(KVPY-2014)

Solution

(c)

Let $\sigma=h^\alpha k^\beta c^\gamma$, then equating dimensions of both sides, we have

$\left[ MT ^{-3} K ^{-4}\right]=\left[ ML ^2 T ^{-1}\right]^\alpha\left[ ML ^2 T ^{-2} K ^{-1}\right]^\beta\left[ LT ^{-1}\right]^\gamma$

$\alpha+\beta=1 \quad…(i)$

$2 \alpha+2 \beta+\gamma=0 \quad \ldots (ii)$

$-\alpha-2 \beta-\gamma=-3 \quad \ldots (iii)$

In Eq. $(i)$, we multiplying with $2$, we get

$2 \alpha+2 \beta=2 \ldots(iv)$

Now, subtracting Eq. $(ii)$ from Eq. $(iv)$,

we get

$2 \alpha-2 \alpha+2 \beta-2 \beta+\gamma=0-2$

$\gamma=-2$

Now, putting the value of $\gamma$ in Eq. $(iii)$, we get

$\alpha+2 \beta=5 \ldots( v )$

And solve the Eq. $(i)$ and $(v)$, we get $\alpha=-3, \beta=4$

So, $\alpha=-3, \beta=4$ and $\gamma=-2$

Standard 11

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The dimensions of Stefan-Boltzmann's constant $\sigma$ can be written in terms of Planck's constant $h$, Boltzmann's constant $k_B$ and the speed of light $c$ as $\sigma=h^\alpha k_B^\beta c^\gamma$. Here,

Solution

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