The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length?

First, we recall that the Planck length $$\ell_P$$ must be built only from the universal constants $$G$$ (Newton’s gravitational constant), $$h$$ (Planck’s constant) and $$c$$ (speed of light). Therefore we suppose $$\ell_P = G^{\alpha}\,h^{\beta}\,c^{\gamma}$$ where $$\alpha,\;\beta,\;\gamma$$ are real numbers to be determined by dimensional analysis.

Now we write the dimensions of each quantity in the fundamental symbols $$M$$ for mass, $$L$$ for length and $$T$$ for time:

$$[G]=M^{-1}L^{3}T^{-2},\qquad [h]=ML^{2}T^{-1},\qquad [c]=LT^{-1}.$$

Hence the dimensions of the assumed product are

$$[G^{\alpha}h^{\beta}c^{\gamma}] = M^{-\,\alpha}L^{3\alpha}T^{-2\alpha}\; M^{\beta}L^{2\beta}T^{-\beta}\; L^{\gamma}T^{-\gamma}.$$

Collecting the powers of each fundamental dimension, we obtain

$$M^{-\,\alpha+\beta}\;L^{\,3\alpha+2\beta+\gamma}\;T^{-\,2\alpha-\beta-\gamma}.$$

The Planck length is, by definition, a quantity of dimension length alone, so its overall dimensions must be $$M^{0}L^{1}T^{0}.$$ Therefore we write three simultaneous equations by equating the exponents:

For mass: $$-\,\alpha+\beta = 0.$$

For length: $$3\alpha + 2\beta + \gamma = 1.$$

For time: $$-\,2\alpha - \beta - \gamma = 0.$$

From the first equation we have $$\beta = \alpha.$$ Substituting this into the time equation gives

$$-\,2\alpha - \alpha - \gamma = 0 \;\;\Rightarrow\;\; -3\alpha - \gamma = 0 \;\;\Rightarrow\;\; \gamma = -3\alpha.$$

Now we use these results in the length equation:

$$3\alpha + 2\alpha + (-3\alpha) = 1 \;\;\Rightarrow\;\; 2\alpha = 1 \;\;\Rightarrow\;\; \alpha = \dfrac{1}{2}.$$

Because $$\beta = \alpha,$$ we have $$\beta = \dfrac{1}{2},$$ and because $$\gamma = -3\alpha,$$ we have $$\gamma = -\dfrac{3}{2}.$$

Therefore

$$\ell_P = G^{\frac{1}{2}}\,h^{\frac{1}{2}}\,c^{-\frac{3}{2}} = \left(\dfrac{G\,h}{c^{3}}\right)^{\frac{1}{2}}.$$

This expression is exactly the same as Option B.

Hence, the correct answer is Option B.