Question

A quantity $\alpha$ is defined as $\alpha = \frac{e^2}{4\pi\epsilon_0 c\hbar}$, where e is electric charge, $\hbar = \frac{h}{2\pi}$ is the reduced Plancks constant and $c$ is the speed of light. The dimensions of $\alpha$ are

• A. $[M^0 L^0 T^0 I^0]$
• B. $[M^1 L^{-1} T^2 I^{-2}]$
• C. $[M^2 L^1 T^{-1} I^0]$
• D. $[M^0 L^3 T^{-1} I^{-2}]$

Answer 1

Answer: (A)

[M^0 L^0 T^0 I^0]

Answer 2

To find the dimensions of the quantity $\alpha = \frac{e^2}{4\pi\epsilon_0 c\hbar}$, we need to determine the dimensions of each component in the expression. The constant $4\pi$ is dimensionless.

The dimensions of the components are:

• Electric charge $e$: $[e] = [I T]$, where $I$ represents electric current and $T$ represents time.
• Speed of light $c$: $[c] = [L T^{-1}]$, where $L$ represents length.
• Reduced Planck's constant $\hbar$: Planck's constant $h$ has the dimensions of energy multiplied by time (action). The dimension of energy is $[M L^2 T^{-2}]$ (from $E = \frac{1}{2}mv^2$). So, $[h] = [M L^2 T^{-2}] \times [T] = [M L^2 T^{-1}]$. Since $\hbar = \frac{h}{2\pi}$ and $2\pi$ is dimensionless, $[\hbar] = [M L^2 T^{-1}]$.
• Permittivity of free space $\epsilon_0$: We can find the dimensions of $\epsilon_0$ from Coulomb's Law, $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$. Rearranging for $\epsilon_0$, we get $\epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. The dimension of force $F$ is $[M L T^{-2}]$. The dimensions of charges $q_1$ and $q_2$ are $[I T]$. The dimension of distance $r$ is $[L]$. So, $[\epsilon_0] = \frac{[I T][I T]}{[M L T^{-2}][L^2]} = \frac{[I^2 T^2]}{[M L^3 T^{-2}]} = [M^{-1} L^{-3} T^4 I^2]$.

Now, substitute these dimensions into the expression for $\alpha$: $[\alpha] = \frac{[e^2]}{[\epsilon_0] [c] [\hbar]}$ $[e^2] = ([I T])^2 = [I^2 T^2]$

Denominator dimensions = $[\epsilon_0] [c] [\hbar]$ $= [M^{-1} L^{-3} T^4 I^2] \times [L T^{-1}] \times [M L^2 T^{-1}]$ Combine the powers of each fundamental dimension: $M$: $M^{-1} \times M^1 = M^{-1+1} = M^0$ $L$: $L^{-3} \times L^1 \times L^2 = L^{-3+1+2} = L^0$ $T$: $T^4 \times T^{-1} \times T^{-1} = T^{4-1-1} = T^2$ $I$: $I^2$ So, the dimensions of the denominator are $[M^0 L^0 T^2 I^2]$.

Now, calculate the dimensions of $\alpha$: $[\alpha] = \frac{[I^2 T^2]}{[M^0 L^0 T^2 I^2]}$ Combine the powers of each fundamental dimension: $M$: $M^0 / M^0 = M^{0-0} = M^0$ $L$: $L^0 / L^0 = L^{0-0} = L^0$ $T$: $T^2 / T^2 = T^{2-2} = T^0$ $I$: $I^2 / I^2 = I^{2-2} = I^0$

Thus, the dimensions of $\alpha$ are $[M^0 L^0 T^0 I^0]$. This quantity $\alpha$ is the fine-structure constant, which is a dimensionless fundamental constant.