In this problem we define the potential $U:\mathbb{R^+}\rightarrow \mathbb{R^+}$ where $\mathbb{R}^+ := (0,+\infty)$ as

$$U(x):= \frac{b}{x^2}-\frac{a}{x} \tag{1}$$

If we suppose that an proton, with initial velocity $v_0$, is coming from infinity to the region subjected to the potential above, we have that, at each $x \in \mathbb{R}^+$ the total energy is

$$E_0 = \frac{mv_0^2}{2} + U(x)$$

**My question 1 is:** *If a photon is emitted at position $x'$ what is the minimum energy of the photon in order to set the proton trapped in the potential well?*

My goal is to note that, if the proton is coming from infinity then it has an energy $E_0$ that it is larger than zero so $E_0 \geq 0$. In order to get the proton in the well we just need that the energy of the proton after the emission will be less then zero $E < 0$. *Is this answer correct?*

The figure above is a plot of the potential $U$ given in $(1)$. The limit of $U$ when $x \rightarrow +\infty$ is zero, so if $E < 0$, at some point, there will be a $x'$ such that the path of the proton will be oscilating between the two points that are the solution of equation $$U(x) = U(x')$$ for $x'$ fixed.

**My question 2 is:** *Can I find the energy of the photon such that in $x'$ gets zero velocity immediately after the emission?*

My goal here is to use (total) energy conservation

$$\frac{mv*0^2}{2} + U(x') = 0 + U(x') + h\nu \Rightarrow h\nu*{min} = \frac{mv_0^2}{2}$$

*Is this correct? If there is an emission, can I use conservation of total mechanical energy* $T+V$?