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

To construct a dimensionless quantity from \( e \), \(\epsilon_0\), \( h \), and \( c \), we assign their respective dimensional formulas with powers \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\). The expression takes the form:

\( e^\alpha \epsilon_0^\beta h^\gamma c^\delta \)

The dimensions of each quantity are:

• \( [e] = [A][T] = [I][T] \)
• \( [\epsilon_0] = [M]^{-1}[L]^{-3}[T]^4[I]^2 \)
• \( [h] = [M][L]^2[T]^{-1} \)
• \( [c] = [L][T]^{-1} \)

The dimensionless condition implies:

\([M]^0[L]^0[T]^0[I]^0\) for the product \( e^\alpha \epsilon_0^\beta h^\gamma c^\delta \)

Equating powers for each dimension:

1. Mass (M):
\(0 = \beta(-1) + \gamma(1) = -\beta + \gamma\) (Equation 1)

2. Length (L):
\(0 = \beta(-3) + \gamma(2) + \delta(1) = -3\beta + 2\gamma + \delta\) (Equation 2)

3. Time (T):
\(0 = \beta(4) + \gamma(-1) + \delta(-1) + \alpha(1) = 4\beta - \gamma - \delta + \alpha\) (Equation 3)

4. Current (I):
\(0 = \alpha(1) + \beta(2) = \alpha + 2\beta\) (Equation 4)

By solving these equations systematically:

From Equation 4: \(\alpha = -2\beta\)

Substitute \(\alpha = -2\beta\) into Equation 3:
\(4\beta - \gamma - \delta - 2\beta = 0\)
\(2\beta - \gamma - \delta = 0\) (Equation 5)

Use Equation 1:\( -\beta + \gamma = 0 \ => \ \gamma = \beta\)

Substitute \(\gamma = \beta\) in Equation 5:
\(2\beta - \beta - \delta = 0\)
\(\beta = \delta\)

Hence, \(\alpha = -2\beta = -2\delta\)

Thus, \(\alpha = 2n\), \(\beta = -n\), \(\gamma = -n\), \(\delta = -n\), as \(n\) is a non-zero integer.

The tuple \((\alpha, \beta, \gamma, \delta)\) is \((2n, -n, -n, -n)\).

Answer 2

Let's verify the correct set of exponents \((\alpha, \beta, \gamma, \delta)\) so that

\[ e^\alpha \epsilon_0^\beta h^\gamma c^\delta \] is dimensionless. ---

Step 1: Dimensions of each quantity

- Charge \(e\): \([I T]\) - Permittivity \(\epsilon_0\): \([M^{-1} L^{-3} T^4 I^2]\) - Planck's constant \(h\): \([M L^2 T^{-1}]\) - Speed of light \(c\): \([L T^{-1}]\) ---

Step 2: Express combined dimension

\[ [I]^{\alpha + 2 \beta} \cdot [M]^{-\beta + \gamma} \cdot [L]^{-3 \beta + 2 \gamma + \delta} \cdot [T]^{\alpha + 4 \beta - \gamma - \delta} \] ---

Step 3: Set exponents to zero

\[ \begin{cases} \alpha + 2 \beta = 0 \\ -\beta + \gamma = 0 \\ -3 \beta + 2 \gamma + \delta = 0 \\ \alpha + 4 \beta - \gamma - \delta = 0 \end{cases} \] ---

Step 4: Solve the system

From second: \(\gamma = \beta\) From first: \(\alpha = -2 \beta\) Substitute in third: \[ -3 \beta + 2 \beta + \delta = 0 \implies -\beta + \delta = 0 \implies \delta = \beta \] Substitute in fourth: \[ -2 \beta + 4 \beta - \beta - \beta = 0 \implies 0=0 \] All consistent. ---

Step 5: Express in terms of \(n = \beta\)

\[ (\alpha, \beta, \gamma, \delta) = (-2n, n, n, n) \] ---

Step 6: Compare with given options

Given options are multiples of \(n\), so multiply both sides by \(-1\) (valid since \(n\) is any non-zero integer): \[ (2n, -n, -n, -n) \] This matches the first option. ---

Final answer:

\[ \boxed{ (2n, -n, -n, -n) } \]