# Does the Bohr model violate the uncertainty principle?

In a book, it is stated that one reason for the failure of Bohr's theory was the fact that it violates the uncertainty principle. Is this fact true? How so?

• There is an issue of semantics here. The Bohr model predates the Uncertainty principle, so one can't really speak to the impropriety 'violation' implies. However, Bohr's model is not compatible with the Uncertainty principle, and the two really can not be rigorously combined to model the quantum mechanics of atoms. – Lighthart Jul 21 '15 at 21:29

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.
In the Bohr model of the atom the energy (thus its momentum) of the electron and the radius of its orbit (thus its position) are precisely defined quantities. This is a direct violation of the Uncertainty Principle. Since the energy levels of the hydrogen atom had been shown to be experimentally well defined by Bohr’s model ($\Delta$p ~ 0), we are left with the conclusion that we know very little about the exact location of the electrons in space ($\Delta$x ~ infinity).