Optics Physics Homework 1 (2016)

1. What is the energy in eV of photons at the ends of the visible spectrum, that is, at wavelenghts of 400 nm and 700 nm?

2. Determine the wavelength and momentum of a photon whose energy equals the rest mass energy of an electron.

3.What is the wavelength and momentum of a photon capable to produce an electron-positron pair?

4. Show that the wavelength of a photon measured in nanometers can be found from its energy measured in electron-volts by the convenient relation

$\lambda = 1240/E(eV)$

Energy of photons is (ν is frequency,λ is wavelength)

$E=hν=hc/λ$

$E_1=(6.626*10^{-34}*3*10^8)/(400*10^{-9} )=4.97*10^{-19} J=$

$=(4.97*10^{-19})/(1.6*10^{-19} ) eV=3.106 eV$

$E_2=(6.626*10^{-34}*3*10^8)/(700*10^{-9} )=2.84*10^{-19} J=$

$=(2.84*10^{-19})/(1.6*10^{-19} )=1.775 eV$

Electron rest mass is $m_0=9.1*10^{-31} kg$

Equivalent energy is $E=m_0 c^2=9.1*10^{-31}*(3*10^8)^2=8.19*10^{-14} J$

Photon energy is

$E_{ph}=hν=hc/λ$ so that

$λ=hc/E_{ph} =(6.626*10^{-34}*3*10^8)/(8.19*10^{-14} )=2.427*10^{-12} m=2.427 pm$

Rest mass of photon is zero it means that its energy is (p is momentum)

$E=pc$ so that

$p=E_ph/c=(8.19*10^{-14})/(3*10^8)=2.73*10^{-22}(kg*m/s)$

$ν→e^-+e^+$ (photon→electron+positron)

Assuming the minimum energy that produces the pair (the electron and positron are at rest) we have

$E_{ph}=2m_0 c^2$ so that $h c/λ=2m_0 c^2$ or

$λ=hc/(2m_0 c^2 )=(6.626*10^{-34}*3*10^8)/(2*9.1*10^{-31}*(3*10^8 )^2)=$

$=(2.427 )/2 pm=1.214 pm$

Momentum of photon is:

$p=E/c=(2m_0 c^2)/c=2*2.73*10^{-22}=5.46*10^{-22} (kg*m/s)$

$E (J)=hν=h c/(λ (m))$ or

$E(eV)*1.6*10^{-19}=(6.626*10^{-34}*3*10^8)/(λ (nm)*10^{-9} )$

$λ (nm)=(6.626*10^{-34}*3*10^8)/(E(eV)*1.6*10^{-19}*10^{-9} )$ or

$λ(nm)=1242/(E(eV))$