Question

As in electron-positron annihilation, when a proton and antiproton annihilate. Twophoton can be produced. Ofwhat wavelength are of these photons?

Short answer

The wavelength of the photon is approximate$\mathbf{1}\mathbf{.3}\mathbf{\text{\hspace{0.17em}fm}}$ .

Step-by-step solution

Given data

The given radiation is the photon.

Concept of Photon energy and Annihilation

The energy of a photon$\mathit{E}$ for some wavelength$\mathit{\lambda }$ is given as:

$E=\frac{hc}{\lambda }$

Here$h$ is Planck's constant and is the speed of light in a vacuum.

Annihilation is the process of combining electron and positron and liberating radiation energy that is equal to the energy of two rest mass photons.

So, the energy is equal to the energy of two photons.

Calculation for the electron-positron energy

The pair of electron-positron will have energy:

$E=2m{c}^2$

Substitute $9.11\times {10}^{-31}\text{kg}$ for the mass of the electron $m$ and $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$ for the speed of me light $c$ in the above equation as:

$\begin{array}{l}E=2(1.67\times {10}^{-27}\text{\hspace{0.17em}kg}){(3\times {10}^8\text{\hspace{0.17em}m/s})}^2\\ E=30.06\times {10}^{-11}\text{\hspace{0.17em}J}\end{array}$

So each photon will have half the energy:

$\begin{array}{l}{E}^{\text{'}}=\frac{E}{2}\\ E^{\text{'}}=\frac{30.06\times {10}^{-11}\text{\hspace{0.17em}J}}{2}\\ E^{\text{'}}=15.03\times {10}^{-11}\text{\hspace{0.17em}J}\end{array}$

 Step 4: Calculation for the wavelength of the photon 

Photon's energy is given as,

$E=\frac{hc}{\lambda }$

Rearrange the above equation for wavelength as,

$\lambda =\frac{hc}{E}$

Substitute $15.03\times {10}^{-11}\text{\hspace{0.17em}J}$ for $E,3\times {10}^8\text{\hspace{0.17em}m/s}$ for speed of the light $c$ and $6.626\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$for h in the above equation as:

$\begin{array}{l}\lambda =\frac{(6.626\times {10}^{-34}\text{J}\cdot \text{s})(3\times {10}^8\text{\hspace{0.17em}m/s})}{(15.03\times {10}^{-11}\text{\hspace{0.17em}J})}\\ \lambda =1.32\times {10}^{-15}\text{m}\left(\frac{1.0\text{\hspace{0.17em}fm}}{1.0\times {10}^{-15}\text{\hspace{0.17em}m}}\right)\\ \lambda =1.3\text{\hspace{0.17em}fm}\end{array}$

The wavelength of the photon is approximately $1.3\text{\hspace{0.17em}fm}$.