Question

The Bohr radius is given by a0=ε0h2πme2. Verify that the RHS has dimensions of length.

Answer 1

Here y $a_0$ = $\varepsilon_{0}$$h^{2}}{\pi m e^{2}}$ is verified that the RHS of Bohr radius has the dimensions of length.

Explanation:

Generally, the Bohr radius is given as “$a_0$ = $\varepsilon_{0}$$h^{2}}{\pi m e^{2}}$”, This was first derived by a Niels Bohr and so it gets this name.

Now consider the right-hand side of the equation $a_0$ we get,

$a=\frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}$

By solving the above equation we get the following derivation,

$\frac{M^{2} L^{2} T^{-2}}{M^{2} L^{3} T^{-2}}=L$

Thus it is now proved that the right hand side of Bohr radius has the dimensions of length.