SEMI FINAL STAGE 2026
Mfantsipim School: 56 points
Presbyterian Boys' Sec. Sch: 53 points
Osei Tutu SHS: 27 points
QUESTION
Give the threshold photon energy, in electron volts, for electron-positron pair production.
ANSWER: $1.022 \times 10^6$ eV
SOLUTION:
$E_{th} = 2m_e c^2$
$m_e c^2 = 0.511 \text{ MeV}$
$E_{th} = 2 \times 0.511 \text{ MeV}$
$E_{th} = 1.022 \text{ MeV}$
$E_{th} = 1.022 \times 10^6 \text{ eV}$
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PRACTICE QUESTIONS
1. QUESTION: A photon of energy 2.02 MeV produces an electron-positron pair. Find the total kinetic energy of the pair.
ANSWER: 0.998 MeV or $9.98\times10^{-1}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 2.02 - 1.022$
$KE_{\text{total}} = 0.998\text{ MeV}$
2. QUESTION: A photon of energy 3.02 MeV produces an electron-positron pair. Find the total kinetic energy of the pair.
ANSWER: 2.00 MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 3.02 - 1.022$
$KE_{\text{total}} = 1.998\text{ MeV}$
3. QUESTION: A photon of energy 4.02 MeV produces an electron-positron pair. Find the total kinetic energy of the pair.
ANSWER: 3.00 MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 4.02 - 1.022$
$KE_{\text{total}} = 2.998\text{ MeV}$
4. QUESTION: A photon of energy 1.62 MeV produces an electron-positron pair. Find the total kinetic energy of the pair.
ANSWER: 0.598 MeV or $5.98\times10^{-1}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 1.62 - 1.022$
$KE_{\text{total}} = 0.598\text{ MeV}$
5. QUESTION: A photon of energy 5.02 MeV produces an electron-positron pair. Find the total kinetic energy of the pair.
ANSWER: 4.00 MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 5.02 - 1.022$
$KE_{\text{total}} = 3.998\text{ MeV}$
6. QUESTION: A photon of energy 1.52 MeV produces an electron-positron pair that shares the energy equally. Find the kinetic energy of each particle.
ANSWER: 0.249 MeV or $2.49\times10^{-1}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 1.52 - 1.022$
$KE_{\text{total}} = 0.498\text{ MeV}$
$KE_{\text{each}} = \dfrac{KE_{\text{total}}}{2}$
$KE_{\text{each}} = \dfrac{0.498}{2}$
$KE_{\text{each}} = \dfrac{498\times10^{-3}}{2}$
$KE_{\text{each}} = 0.249\text{ MeV}$
7. QUESTION: A photon of energy 1.42 MeV produces an electron-positron pair that shares the energy equally. Find the kinetic energy of each particle.
ANSWER: 0.199 MeV or $1.99\times10^{-1}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 1.42 - 1.022$
$KE_{\text{total}} = 0.398\text{ MeV}$
$KE_{\text{each}} = \dfrac{KE_{\text{total}}}{2}$
$KE_{\text{each}} = \dfrac{0.398}{2}$
$KE_{\text{each}} = \dfrac{398\times10^{-3}}{2}$
$KE_{\text{each}} = 0.199\text{ MeV}$
8. QUESTION: A photon of energy 1.82 MeV produces an electron-positron pair that shares the energy equally. Find the kinetic energy of each particle.
ANSWER: 0.399 MeV or $3.99\times10^{-1}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 1.82 - 1.022$
$KE_{\text{total}} = 0.798\text{ MeV}$
$KE_{\text{each}} = \dfrac{KE_{\text{total}}}{2}$
$KE_{\text{each}} = \dfrac{0.798}{2}$
$KE_{\text{each}} = \dfrac{798\times10^{-3}}{2}$
$KE_{\text{each}} = 0.399\text{ MeV}$
9. QUESTION: A photon of energy 1.22 MeV produces an electron-positron pair that shares the energy equally. Find the kinetic energy of each particle.
ANSWER: 0.0990 MeV or $9.90\times10^{-2}$ MeV
SOLUTION:
$E_{th} = 2m_ec^2$
Memorise: $E_{th} = 1.022\text{ MeV}$ (twice the electron's rest energy of 0.511 MeV)
$KE_{\text{total}} = E - E_{th}$
$KE_{\text{total}} = 1.22 - 1.022$
$KE_{\text{total}} = 0.198\text{ MeV}$
$KE_{\text{each}} = \dfrac{KE_{\text{total}}}{2}$
$KE_{\text{each}} = \dfrac{0.198}{2}$
$KE_{\text{each}} = \dfrac{198\times10^{-3}}{2}$
$KE_{\text{each}} = 0.0990\text{ MeV}$