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Step-by-Step Solution
Step 1: Express the Planck length using a general dimensional form
Assume the Planck length $ \ell $ can be written as a product of the fundamental physical constants raised to some powers:
$ \ell = k \, G^{p} \, \hbar^{q} \, c^{r}
$
where:
$G$ is the gravitational constant.
$\hbar$ is the reduced Planck's constant.
$c$ is the speed of light in vacuum.
$k$ is a dimensionless constant (often taken as 1 in dimensional analysis).
$p$, $q$, and $r$ are the exponents to be determined.
Step 2: Identify the dimensions of each constant
The dimensions in mass ($M$), length ($L$), and time ($T$) of each constant are:
$G \rightarrow [M^{-1} L^3 T^{-2}]$
$\hbar \rightarrow [M\,L^2\,T^{-1}]$
$c \rightarrow [L\,T^{-1}]$
The Planck length $ \ell $ has dimensions of length, which is $[M^0 L^1 T^0]$.
Step 3: Write the combined dimensions
When we combine these constants, the combined dimensions of $G^{p} \hbar^{q} c^{r}$ become:
$ \bigl[M^{-1}L^{3}T^{-2}\bigr]^{p} \times \bigl[M^{1}L^{2}T^{-1}\bigr]^{q} \times \bigl[L\,T^{-1}\bigr]^{r}.
$
Collecting exponents for $M$, $L$, and $T$ gives:
$ M^{-p + q} \,
L^{(3p + 2q + r)} \,
T^{-(2p + q + r)}.
$
Step 4: Equate the dimensions to those of length
We want the final dimensional form to be $[M^{0}L^{1}T^{0}]$. Hence, we set up the system of equations:
For $M$: $-p + q = 0$
For $L$: $3p + 2q + r = 1$
For $T$: $-(2p + q + r) = 0$ or $2p + q + r = 0$
Step 5: Solve for the exponents $p$, $q$, and $r$
Select each equation and solve step by step:
From $-p + q = 0$, we get $q = p$.
From $2p + q + r = 0$, substitute $q = p$ to get $2p + p + r = 0 \implies 3p + r = 0 \implies r = -3p$.
From $3p + 2q + r = 1$, substitute $q = p$ and $r = -3p$:
$3p + 2p + (-3p) = 1 \implies 2p = 1 \implies p = \frac{1}{2}.$
Hence $p = \frac{1}{2}$, $q = \frac{1}{2}$, and $r = -\frac{3}{2}$.
Step 6: Write the final expression for the Planck length
Substitute $p = \tfrac{1}{2}$, $q = \tfrac{1}{2}$, and $r = -\tfrac{3}{2}$ into
$ G^p \hbar^q c^r $:
$ \ell = k \, G^{\tfrac{1}{2}} \,\hbar^{\tfrac{1}{2}} \,c^{-\tfrac{3}{2}}
= k \left(\frac{G \hbar}{c^3}\right)^{\tfrac{1}{2}}.
$
Assuming $k = 1$, we obtain the well-known Planck length:
$ \ell_\text{Planck}
= \left(\frac{G \,\hbar}{c^3}\right)^{\tfrac{1}{2}}.
$
Final Answer
Therefore, the correct form of the Planck length is:
$ \ell_\text{Planck}
= {\left(\frac{G \,\hbar}{c^3}\right)^{\frac{1}{2}}}.
$
