Respuesta :
The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
If we consider the potential energy (U) between the pions, then (U) can be expressed as:
[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]
Given that at some instance, the potential energy becomes negligible compared to the final K.E.
As such the conservation of the total energy in the system can be given as:
- E = U + K
Again, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:
[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]
∴
Equating both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]
[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]
where:
- r = distance
- k = Columb's constant
- q = charge on a proton
- m_o = rest mass of each pion in the previous question
- c = velocity of light
- [tex]\mathbf{v_\pi}[/tex] = calculated velocity of proton in the previous question
Replacing their values in the above equation, the distance (r) between the pions is calculated as:
[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]
distance (r) = 1.45 × 10⁻¹⁶ m
Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
Learn more about electric potential energy here:
https://brainly.com/question/21808222?referrer=searchResults