Does the Bohr model violate the uncertainty principle?

Question

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?

Answer 1

The Heisenberg uncertainty principle states that

$$\Delta x \Delta p \geq \frac{\hbar}{2}$$

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.

The Bohr model was derived via the classical assumption that the electron orbits the nucleus, much like how the Earth orbits the Sun, except that the origin of the centripetal force was electromagnetic instead of gravitational. It then went on to derive the energy levels of the electron in a hydrogen atom using the laws of uniform circular motion and electrostatics. This means that it is theoretically possible to know the position and momentum of the electron simultaneously with no uncertainty - a violation of the uncertainty principle.

Answer 2

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).

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