Evaluate option A

Heisenberg's uncertainty principle states that the exact position and exact momentum of an electron cannot be known simultaneously: $\Delta x\, \Delta p \ge \frac{h}{4\pi}$ Therefore, the idea of a definite classical path for an electron is not valid.

Hence, A is correct.

Evaluate option B

For an electron bound in an atom, its energy is negative relative to an electron at infinite distance, whose energy is taken as zero.

Thus, for an electron in the $2s$ orbital: $E_{2s} < 0$ while at infinite distance: $E_{\infty} = 0$ Therefore, $E_{2s} < E_{\infty}$

Hence, B is correct.

Evaluate option C

In Bohr's model, the energy of an electron in the $n$th orbit is $E_n = -\frac{13.6 Z^2}{n^2}\; \text{eV}$ The most negative value occurs for the smallest possible value of $n$, i.e. $n=1$.

Since lower (more negative) energy means greater stability, $n=1$ is the most stable orbit.

Hence, C is correct.

Evaluate option D

In Bohr's model, electron speed in the $n$th orbit is $v_n = \frac{Z\alpha c}{n}$ so $v_n \propto \frac{1}{n}$ Therefore, as $n$ increases, velocity decreases, not increases.

Hence, D is incorrect.

Final answer

Correct statements are: $\boxed{A,\ B,\ C}$

Comparison with stored correct answer

Stored correct answer: A, B, C

My derived answer matches the stored answer.