Question

The dimensional formula of a physical quantity represented by \( \frac{e^2}{4 \varepsilon_0 h} \) is:
Where \( e \) is the charge of the electron, \( \varepsilon_0 \) is the permittivity of free space, and \( h \) is Planck's constant.

• A. \( [M^1 L^1 T^{-1}] \)
• B. \( [L T^{-1}] \)
• C. \( [M^1 T^{-1}] \)
• D. \( [M^1 L^1 T^{-2}] \)

Hint:

Use base dimensional formulas and simplify algebraically to derive compound expressions.

Answer 1

The Correct Option is B

The dimensional formula for charge \( e = [A T] \), permittivity \( \varepsilon_0 = [M^{-1} L^{-3} T^4 A^2] \), and Planck’s constant \( h = [M L^2 T^{-1}] \).
So, \[ \frac{e^2}{\varepsilon_0 h} = \frac{[A^2 T^2]}{[M^{-1} L^{-3} T^4 A^2] [M L^2 T^{-1}]} = \frac{1}{L^{-1} T^1} = [L T^{-1}] \]