In an accelerator experiment on high-energy c | vedclass

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Question

(C) Total energy of the electron-positron pair is $E = 10.2 \; BeV = 10.2 	imes 10^9 \; eV$.Converting this energy into Joules: $E = 10.2 	imes 10^9

Vedclass · Accepted Answer

$2.436 	imes 10^{-16} \; m$

Vedclass · Answer

(C) Total energy of the electron-positron pair is $E = 10.2 \; BeV = 10.2 	imes 10^9 \; eV$.Converting this energy into Joules: $E = 10.2 	imes 10^9 	imes 1.602 	imes 10^{-19} \; J = 1.634 	imes 10^{-9} \; J$.Since the pair annihilates into two $\gamma$-rays of equal energy,the energy of each $\gamma$-ray is $E' = E / 2 = 0.817 	imes 10^{-9} \; J$.Using the relation $E' = \frac{hc}{\lambda}$,the wavelength $\lambda$ is given by $\lambda = \frac{hc}{E'}$.Substituting $h = 6.626 	imes 10^{-34} \; Js$ and $c = 3 	imes 10^8 \; m/s$:$\lambda = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{0.817 	imes 10^{-9}} \approx 2.433 	imes 10^{-16} \; m$.Rounding to the provided option,the correct wavelength is $2.436 	imes 10^{-16} \; m$.

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