(f) As the pions move away from each other, the ratio of the absolute value of electric potential energy to the final total kine

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(f) As the pions move away from each other, the ratio of the absolute value of electric potential energy to the final total kine

tic energy of the two pions changes. At some point, the potential energy becomes negligible compared to the final total kinetic energy . We can consider that the value of the ratio is about 0.01 when , where is the final total kinetic energy of the two pions. What will the distance, , between the pions be when this criterion is satisfie

Physics

1 answer:

Scorpion4ik [409] · Answer 1 · 2022-07-15T06:14:07+03:00

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:

$\mathbf{U = \dfrac{kq^2}{r} ---- (1)}$

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:

$\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}$

∴

Equating both equations (1) and (2) together, we have:

$\mathbf{\dfrac{kq^2}{r} = 0.01 K}$

$\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }$

$\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}$

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
$\mathbf{v_\pi}$ = calculated velocity of proton in the previous question

Replacing their values in the above equation, the distance (r) between the pions is calculated as:

$\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}$