The positrons near the galactic center can be detected because they annihilate with electrons and emit gamma rays. Two gamma rays are produced in such a situation to conserve both energy and momentum. Suppose the positron and electron, each with energy mc², convert all of this energy into two identical gamma-ray photons. What will be the wavelength of these photons? The mass of an electron is 9.1 * 10^-31 kg.

Question

Quizplus · Accepted Answer

The energy of each photon is given by $E = mc^2$, where $m$ is the mass of the electron (or positron) and $c$ is the speed of light. Since both the electron and positron contribute their mass energy to the two photons, the energy of one photon is $E = mc^2$. The wavelength $\lambda$ of the photon can be found using the equation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg} / \text{s}$). Given the mass of an electron is $9.1 \times 10^{-31}$ kg, and using $c = 3.00 \times 10^8 \, \text{m/s}$, we can calculate the energy of one photon and then its wavelength. The calculation yields a wavelength in the order of $10^{-12}$ meters, which corresponds to the option that indicates a wavelength in nanometers (nm) in the order of $10^{-12}$ meters, which is option B, $2.4 \times 10^{-12}$ meters or $2.4 \times 10^{-3}$ nm.

The Positrons near the Galactic Center Can Be Detected Because