Question

What are the most significant differences between the Bohr model of the hydrogen atom and the Schrödinger analysis? What are the similarities?

Short answer

Similarities, Both predict the energy levels of the hydrogen atom by $E_n=\frac{-13.6eV}{n^2}$.

Differences, In Bohr's model, the electron is a particle moving in a uniform circle within one dimension (radial), so it has only one quantum number, and its angular momentum is , so for the ground state, it is . In Schrodinger's model, the electron is a cloud of probability amplitude in three-dimensional motion, with its motion described by a wave function defined by three quantum numbers. And the angular momentum is $L=ħ\sqrt{l\left(l+1\right)}$, so for the ground state, it is zero.

Step-by-step solution

Similarities

Both models predict the energy levels of the hydrogen atom correctly that are given by$E_n=\frac{-13.6eV}{n^2}$ with n = 1,2, 3,….

Differences

a)

Bohr's model pictures the electron as a particle.

Schrodinger's model pictures the electron as a cloud of probability amplitude with its motion described by a wave equation.

b)

In Bohr's model, the electron is moving in a flat circle in one-dimensional motion (radial) like a classical particle described by Newton's laws, so it has only one quantum number, n;

In Schrodinger's model, the electron is moving in the three-dimensional space around the hydrogen nucleus, r, $\theta$, $\varphi$, so a state is defined by three quantum numbers, n,l ,$m_l$ .

c)

In the Bohr model, angular momentum can take the values $L=nħ$n =1,2,3, so the ground-state angular momentum is $ħ$.

In the Schrodinger model, angular momentum can take the values$L=ħ\sqrt{l\left(l+1\right)}$.

role="math" localid="1663917003433" $l$= 0,1,... , n-1, so the ground-state angular momentum is zero.