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QPages 384-386: Q1: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q1 Solution: The Bohr radius is given by . Verify that the RHS has…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently misremember the dimensions of the permittivity of free space ($\varepsilon_0$) as $[M^{-1} L^{-3} T^4 I^2]$ or confuse it with the permeability ($\mu_0$), leading to incorrect mass or current exponents. Always derive or verify $[\varepsilon_0]$ using the force law $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$ to ensure the final result simplifies correctly to $[L]$.

The Bohr radius is given by a0=ε0h2πme2a_0 = \frac{\varepsilon_0 h^2}{\pi m e^2}. Verify that the RHS has dimensions of length.

Step-by-Step Verified Solution

**Answer:** The RHS has dimensions of length. **Step-by-Step Explanation:**
a0=ϵ0h2πme2a_0 = \frac{\epsilon_0 h^2}{\pi m e^2}
Dimensions: - [ϵ0]=[M−1L−3T4I2][\epsilon_0] = [M^{-1} L^{-3} T^4 I^2] - [h]=[ML2T−1][h] = [M L^2 T^{-1}] - [m]=[M][m] = [M] - [e]=[IT][e] = [I T] RHS dimensions:
[M−1L−3T4I2]×[M2L4T−2][M]×[I2T2]=[MLT2I2][MI2T2]=[L]\frac{[M^{-1} L^{-3} T^4 I^2] \times [M^2 L^4 T^{-2}]}{[M] \times [I^2 T^2]} = \frac{[M L T^2 I^2]}{[M I^2 T^2]} = [L]
Hence, the RHS has dimensions of length [citation:3][citation:11].

All Questions in this Chapter

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