Question 262: Units , Dimensions and Error Analysis

Question

A dimensionless quantity is constructed in terms of electronic charge $e$, permittivity of free space $\varepsilon_0$, Planck's constant $h$, and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer, then $(\alpha, \beta, \gamma, \delta)$ is given by :

Accepted Answer

To find the values of $(\alpha, \beta, \gamma, \delta)$ for which the expression $e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta$ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge ($e$), permittivity ($\varepsilon_0$), Planck's constant ($h$), and speed of light ($c$) need to cancel each other out.
Let's look at the dimensions of each of the physical quantities:
Electronic charge ($e$):
The dimension of charge in terms of fundamental dimensions is:
$$[e] = [A \cdot T]$$
where $A$ is the dimension of current and $T$ is the dimension of time.
Permittivity of free space ($\varepsilon_0$):
The dimension is given by:
$$[\varepsilon_0] = [	ext{M}^{-1} 	ext{L}^{-3} 	ext{T}^4 	ext{A}^2]$$
Planck's constant ($h$):
The dimension is:
$$[h] = [	ext{M} 	ext{L}^2 	ext{T}^{-1}]$$
Speed of light ($c$):
The dimension is:
$$[c] = [	ext{L} 	ext{T}^{-1}]$$
Now, we write the expression for the dimensionless quantity and set its dimensions to zero:
$$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [A^\alpha T^\alpha] [	ext{M}^{-\beta} 	ext{L}^{-3\beta} 	ext{T}^{4\beta} 	ext{A}^{2\beta}] [	ext{M}^\gamma 	ext{L}^{2\gamma} 	ext{T}^{-\gamma}] [	ext{L}^\delta 	ext{T}^{-\delta}]$$
Combining the dimensions, we get:
$$[e^\alpha \varepsilon_0{ }^\beta h^\gamma c^\delta] = [	ext{M}^{-\beta + \gamma} 	ext{L}^{-3\beta + 2\gamma + \delta} 	ext{T}^{\alpha + 4\beta - \gamma - \delta} 	ext{A}^{\alpha + 2\beta}]$$
For the quantity to be dimensionless, the exponents of $	ext{M}$, $	ext{L}$, $	ext{T}$, and $	ext{A}$ must all be zero:
$$-\beta + \gamma = 0$$
$$-3\beta + 2\gamma + \delta = 0$$
$$\alpha + 4\beta - \gamma - \delta = 0$$
$$\alpha + 2\beta = 0$$
Solving the above set of equations, we start with the fourth equation:
$$\alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$$
Substituting $\alpha = -2\beta$ into the third equation, we get:
$$-2\beta + 4\beta - \gamma - \delta = 0 \Rightarrow 2\beta - \gamma - \delta = 0 \Rightarrow \gamma + \delta = 2\beta$$
From the first equation:
$$-\beta + \gamma = 0 \Rightarrow \gamma = \beta$$
Then, substituting $\gamma = \beta$ into $\gamma + \delta = 2\beta$:
$$\beta + \delta = 2\beta \Rightarrow \delta = \beta$$
Thus, we have:
$$\alpha = -2\beta$$
$$\gamma = \beta$$
$$\delta = \beta$$
Let $\beta = -n$ where $n$ is a non-zero integer. Then, we can write the values as:
$$\alpha = -2(-n) = 2n$$
$$\beta = -n$$
$$\gamma = -n$$
$$\delta = -n$$
Hence, the tuple $(\alpha, \beta, \gamma, \delta)$ is given by:
$$ (2n, -n, -n, -n) $$
So, the correct option is
Option A:
$(2n,-n,-n,-n)$

Answer

$(2 n,-n,-n,-n)$

Answer

$(n,-n,-2 n,-n)$

Answer

$(n,-n,-n,-2 n)$

Answer

$(2 n,-n,-2 n,-2 n)$

Units , Dimensions and Error Analysis

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