The Question is from neet-physics units-and-measurement conversion-of-units-for-one-system-to-another, and it was created by the author VENKATESWARARAO

Question

The dimensional formula of $\frac{1}{{{\varepsilon _0}}}\frac{{{e^2}}}{{hc}}$ is

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Dimensional Analysis
	**Related Topics**: Fundamental constants, Electromagnetic theory
	**Prerequisites**: Knowledge of dimensions of fundamental physical quantities, basic electromagnetic concepts.
## SOLUTION PATHWAY
**Given Information**:
	Expression: $\frac{1}{{{\varepsilon _0}}}\frac{{{e^2}}}{{hc}}$
	Where:
	  - $\varepsilon_0$ is the permittivity of free space
	  - $e$ is the elementary charge
	  - $h$ is Planck's constant
	  - $c$ is the speed of light

**Method & Steps**:
	1. Identify the dimensions of each quantity:
		- Dimension of $\varepsilon_0$: From Coulomb's Law $F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2}$, we have $\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$. Thus, the dimensions of $\varepsilon_0$ are $\frac{A^2T^2}{MLT^{-2}L^2} = M^{-1}L^{-3}T^4A^2$  
		- Dimension of $e$: Elementary charge has dimensions of $A T$.
		- Dimension of $h$: From energy $E = h
u$, we have $h = \frac{E}{
u}$. Thus, the dimensions of $h$ are $\frac{ML^2T^{-2}}{T^{-1}} = ML^2T^{-1}$.
		- Dimension of $c$: Speed of light has dimensions of $LT^{-1}$.
	2. Substitute the dimensions into the given expression:
		$\frac{1}{\varepsilon_0} \frac{e^2}{hc} = \frac{1}{M^{-1}L^{-3}T^4A^2} \frac{(AT)^2}{ML^2T^{-1}LT^{-1}} = M^1L^3T^{-4}A^{-2} \frac{A^2T^2}{ML^3T^{-2}} = M^0L^0T^0A^0$
	3. Simplify the expression:
		The resulting dimensions are $[M^0L^0T^0A^0]$

**Result**:
	The final answer is $\left[ {{{m{M}}^0}{{m{L}}^0}{{m{T}}^0}{{m{A}}^0}} ight]$
	
 ## OPTION ANALYSIS
[A] ✓: Correct. The dimensions cancel out to yield a dimensionless quantity.
[B] :x: Incorrect. This option does not match the derived dimensions.
[C] :x: Incorrect. This option does not match the derived dimensions.
[D] :x: Incorrect. This option does not match the derived dimensions.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly determining the dimensions of fundamental constants. : Refer to standard tables of dimensional formulas.
		2. Making errors when simplifying the powers of the dimensions. : Double-check the exponents during multiplication and division.
	**Quick Tips**:
		1. Remember that dimensionless quantities often appear in physics, especially in ratios of similar parameters. 
		2. Use the fundamental laws to derive the dimensions of the constants (e.g., Coulomb's Law for $\epsilon_0$, energy of a photon for $h$).

VENKATESWARARAO · Answer

The correct Option is "$\left[ {{{\rm{M}}^0}{{\rm{L}}^0}{{\rm{T}}^0}{{\rm{A}}^0}} \right]$"

VENKATESWARARAO · Answer

$\left[ {{{\rm{M}}^{ - 1}}{{\rm{L}}^3}{{\rm{T}}^2}{\rm{A}}} \right]$

VENKATESWARARAO · Answer

$\left[ {{\rm{M}}{{\rm{L}}^3}{{\rm{T}}^{ - 4}}{{\rm{A}}^{ - 2}}} \right]$

VENKATESWARARAO · Answer

$\left[ {{{\rm{M}}^{ - 1}}{{\rm{L}}^{ - 3}}{{\rm{T}}^4}} \right]$

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