Question

A dimensionless quantity is constructed in terms of electronic charge \(e\), permittivity of free space \(\epsilon_0\) , Planck’s constant ℎ, and speed of light c. If the dimensionless quantity is written as \(e^\alpha\epsilon_0^\beta h^\gamma c^\delta\)and n is a non-zero integer, then\((\alpha, \beta,\gamma,\delta)\) is given by

• A. \((2n,-n,-n,-n)\)
• B. \((n,-n,-2n,-n)\)
• C. \((n,-n,-n,-2n)\)
• D. \((2n,-n,-2n,-2n)\)

Answer 1

The Correct Option is A

Dimensional Analysis 

The given dimensionless quantity is written as:

\[ e^{\alpha} \varepsilon_0^{\beta} h^{\gamma} c^{\delta} \]

The dimensional formula for each term is:

• \( e \rightarrow [T A] \)
• \( \varepsilon_0 \rightarrow [M^{-1}L^{-3}T^4A^2] \)
• \( h \rightarrow [ML^2T^{-1}] \)
• \( c \rightarrow [LT^{-1}] \)

Substituting their powers into the dimensionless quantity:

\[ [T A]^{\alpha} \cdot [M^{-1}L^{-3}T^4A^2]^{\beta} \cdot [ML^2T^{-1}]^{\gamma} \cdot [LT^{-1}]^{\delta} = [L^0 M^0 T^0 A^0] \]

Expanding dimensions:

\[ M^{- \beta + \gamma} L^{-3 \beta + 2\gamma + \delta} T^{\alpha + 4\beta - \gamma - \delta} A^{\alpha + 2\beta} = M^0 L^0 T^0 A^0 \]

Equating powers of \( M, L, T, A \):

• \( -\beta + \gamma = 0 \)   (1)
• \( -3\beta + 2\gamma + \delta = 0 \)   (2)
• \( \alpha + 4\beta - \gamma - \delta = 0 \)   (3)
• \( \alpha + 2\beta = 0 \)   (4)

From (4):

\( \alpha = -2\beta \)   (5)

From (1):

\( \beta = \gamma \)   (6)

Substituting \( \beta = \gamma \) into (2):

\( -3\beta + 2\beta + \delta = 0 \Rightarrow -\beta + \delta = 0 \Rightarrow \delta = \beta \)   (7)

Substituting \( \beta = \gamma = \delta \) and \( \alpha = -2\beta \) into (3):

\( -2\beta + 4\beta - \beta - \beta = 0 \)

This equation is satisfied, so the solution is consistent. Let \( \beta = -n \), where \( n \) is a non-zero integer.

Then:

• \( \alpha = 2n \)
• \( \beta = -n \)
• \( \gamma = -n \)
• \( \delta = -n \)

Final Answer:

\( (2n, -n, -n, -n) \)