91Ó°ÊÓ

First, let's find the total energy of the combined electron and positron before their annihilation. Since we know that the mass of an electron (and that of a positron) is approximately \(9.1 \times 10^{-31}\) kg, according to the Einstein's equation of mass-energy equivalence, we have: \[E = mc^2\] Where \(E\) is the energy, \(m\) is the mass of both particles, and \(c\) is the speed of light (\(3 \times 10^8\) m/s). Here, the mass will be the combined mass of both electron and positron: \[m = 2 \times (9.1 \times 10^{-31}) kg \] Now, calculate the total energy: \[E = 2 \times (9.1 \times 10^{-31}) kg \times (3 \times 10^8 m/s)^2\] As the total energy is divided into two equal-energy photons, we will calculate the energy of one photon: \[E_{photon} = \frac{1}{2}E\]

Now that we have the energy of one photon, we will use the Planck's formula to calculate the wavelength of the photon. The formula is given by: \[E_{photon} = h \times f\] Where \(E_{photon}\) is the energy of the photon, \(h\) is the Planck's constant (\(6.63 \times 10^{-34} Js\)), and \(f\) is the frequency of the photon. We can also write the equation in terms of wavelength using the relationship between frequency and wavelength: \[f = \frac{c}{\lambda}\] Where \(f\) is the frequency, \(c\) is the speed of light and \(\lambda\) is the wavelength. Putting this in the Planck's formula: \[E_{photon} = h \times \frac{c}{\lambda}\] Now, we will solve for the wavelength: \[\lambda = \frac{h \times c}{E_{photon}}\] Plug in the values and calculate the wavelength of the photon.

To determine if the photons are gamma-ray photons, we need to compare their wavelengths with the range of wavelengths typical for gamma rays. Gamma rays usually have wavelengths less than \( 10^{-11} m \). Compare the calculated wavelength of the photons with the typical gamma-ray wavelength range, and conclude whether the photons produced are gamma-ray photons or not.