Skip to main content
deleted 2 characters in body
Source Link

Usually when transforming into dimensionless variables one looks at the relevant constants in the problem. For hydrogen atom we have the electron charge $e$, electron mass $m_e$, Plank constant $\hbar$, permittivity of free space $\epsilon_0$

Then one does dimensional analysis to make scales based on the above constants. And the expression for length scale turns out to be $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}^2}$$$$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}}$$ The extra $4\pi$ is part of the permittivity. And this is exactly the Bohr radius.

Usually when transforming into dimensionless variables one looks at the relevant constants in the problem. For hydrogen atom we have the electron charge $e$, electron mass $m_e$, Plank constant $\hbar$, permittivity of free space $\epsilon_0$

Then one does dimensional analysis to make scales based on the above constants. And the expression for length scale turns out to be $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}^2}$$ The extra $4\pi$ is part of the permittivity. And this is exactly the Bohr radius.

Usually when transforming into dimensionless variables one looks at the relevant constants in the problem. For hydrogen atom we have the electron charge $e$, electron mass $m_e$, Plank constant $\hbar$, permittivity of free space $\epsilon_0$

Then one does dimensional analysis to make scales based on the above constants. And the expression for length scale turns out to be $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}}$$ The extra $4\pi$ is part of the permittivity. And this is exactly the Bohr radius.

Source Link

Usually when transforming into dimensionless variables one looks at the relevant constants in the problem. For hydrogen atom we have the electron charge $e$, electron mass $m_e$, Plank constant $\hbar$, permittivity of free space $\epsilon_0$

Then one does dimensional analysis to make scales based on the above constants. And the expression for length scale turns out to be $$a_0=\frac{4\pi\epsilon_0\hbar^2}{e^2 {m_e}^2}$$ The extra $4\pi$ is part of the permittivity. And this is exactly the Bohr radius.