Question

Consider a process in which two gamma-ray photons are released when an electron and a positron are annihilated. [Note that the mass of a positron is equal to that of an electron, and the gamma-ray photon has no mass.]
(a) What is the final energy of each photon (in J units)?
(b) What is the final energy of each photon (in kJ/mol units)?
(c) What is the final energy of each photon (in eV units)?

Answer 1

Answer:

$\text{(a) }8.187 \times 10^{-14}\text{ J}; \text{(b) 49 300 kJ/mol; (c) } 5.110 \times 10^{5}\text{ eV}$

Explanation:

This question involves the conversion of mass into energy: E = mc².

e + p ⟶ γ + γ

Each particle has the same mass so, in terms of mass, we can write

2e ⟶ 2γ or

 e ⟶ γ

Thus, we can just convert the mass of an electron to its energy equivalent.

(a) Energy in joules

$E= 9.109 \times 10^{-28}\text{ kg} \times \left (2.998 \times 10^{8}\text{ m \cdot s}^{-1}\right )^{2} = \mathbf{8.187 \times 10^{-14}}\textbf{ J}$

(b) Energy in kilojoules per mole

$E =8.187 \times 10^{-14}\text{ J } \times \dfrac{\text{ 1 k}}{\text{1000 J}} \,\times \,\dfrac{6.022 \times 10^{23}}{\text{1 mol}} = 4.930 \times 10^{7}\text{ J/mol}= \textbf{49 300 kJ/mol}$

(c) Energy in electron volts

$E = 8.187 \times 10^{-14}\text{ J}\times \dfrac{6.242 \times 10^{18}\text{ eV}}{\text{1 J}} = \mathbf{5.110 \times 10^{5}}\textbf{ eV}$