# Study of the potential $U(x) = \frac{-a}{x}+\frac{b}{x²}$

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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

$$\frac{mv0^2}{2} + U(x') = 0 + U(x') + h\nu \Rightarrow h\nu{min} = \frac{mv_0^2}{2}$$

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

What is $R^+ \rightarrow R^+$  ?
1. The emission of the photon must reduce the kinetic energy of the proton by $\frac12 mv_0^2$. This is the same for all positions of the proton.