Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that:
$$
b = \sqrt{\frac{\hbar}{m \omega}}
$$
Where:
- $ \hbar $ is the reduced Planck's constant with units of $ \rm{J} \cdot \rm{s} = \rm{kg} \cdot \rm{m}^2 / \rm{s}$
- $ m $ is the mass of the particle in kilograms ($ \rm{kg} $),
- $ \omega $ is the angular frequency with units of $ \rm{s}^{-1} $ (inverse seconds).
So the units of $ b $ are
$$
\text{Units of}\;(b) = \sqrt{\frac{\rm{J} \cdot \rm{s}}{\rm{kg} \cdot \rm{s}^{-1}}} = \sqrt{\frac{\rm{kg} \cdot \rm{m}^2 / \rm{s}}{\rm{kg} / \rm{s}}} = \sqrt{\rm{m}^2} = \boxed{\rm{m}}
$$
$$\color{blue}{\bf [a]}$$
The classical turning point for a harmonic oscillator is the position where the potential energy $ U $ equals the total energy $ E $, meaning the kinetic energy $ K = 0 $. For a harmonic oscillator, the potential energy is:
$$
U(x) = \frac{1}{2} k x^2
$$
At the classical turning point, the total energy $ E $ is equal to the potential energy.
$$
E =U(x) = \frac{1}{2} k x_{\rm{tp}}^2\tag 1
$$
For the quantum harmonic oscillator in the $ n = 1 $ state, the total energy is.
$$
E = \frac{1}{2} \hbar \omega\tag 2
$$
At the turning point, the potential energy is equal to the total energy.
From (1) and (2);
$$
\frac{1}{2} k x_{\rm{tp}}^2 = \frac{1}{2} \hbar \omega
$$
Canceling out the $ \frac{1}{2} $ on both sides;
$$
k x_{\rm{tp}}^2 = \hbar \omega
$$
Solving for $ x_{\rm{tp}} $:
$$
x_{\rm{tp}} = \pm \sqrt{\frac{\hbar \omega}{k}}
$$
Recalling that $ k = m \omega^2$. Substituting this into the equation for $ x_{\rm{tp}} $.
$$
x_{\rm{tp}} = \pm \sqrt{\frac{\hbar \omega}{m \omega^2}} = \pm \sqrt{\frac{\hbar}{m \omega}}
$$
But this expression is just the definition of $ b $, so
$$
\boxed{x_{\rm{tp}} = \pm\; b}
$$
