A $5 \times 10^{- 4} o A$ photon produces an electron-positron pair in the vicinity of a heavy nucleus. Rest energy of electron is 0.5 11 MeV. If they have the same kinetic energies, the energy of each particle is nearly

1.2 MeV

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12 MeV

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120 MeV

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1200 MeV

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Solution

The correct option is B 12 MeV
$\begin{matrix} In the kinetic energy of each particle is k, then m_{e} = m_{e} + 0.511 M e v Energy of photon = \frac{h c}{λ} \end{matrix}$
$\begin{matrix} Applying conservation of energy 2 k + 2 (0.511 M e v) = \frac{h c}{λ} \end{matrix}$
$2 k + 1.022 M e v = \frac{12.4 \times 10^{- 3} M e v A^{\circ}}{5 \times 10^{- 4} A^{0}}$
$2 k + 1.022 M e v = 24.8 M e v$
$K = 11.9 M e v \approx 12 M e V$

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A 5×10−4oA photon produces an electron-positron pair in the vicinity of a heavy nucleus. Rest energy of electron is 0.5 11 MeV. If they have the same kinetic energies, the energy of each particle is nearly

The correct option is B 12 MeV In the kinetic energy of each particle is k , then me=me+0.511Mev Energy of photon = hcλ Applying conservation of energy 2k+2(0.511Mev)=hcλ2k+1.022Mev=12.4×10−3MevA∘5×10−4A02k+1.022Mev=24.8MevK=11.9Mev≈12MeV

A $5 \times 10^{- 4} o A$ photon produces an electron-positron pair in the vicinity of a heavy nucleus. Rest energy of electron is 0.5 11 MeV. If they have the same kinetic energies, the energy of each particle is nearly