In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.

Respuesta :

Answer:

E = 1.24MeV

Explanation:

The photon travels at the speed of light, 3.0 × [tex]10^{8}[/tex] m/s, and given that its frequency = 1 picometer = 1.0 × [tex]10^{-12}[/tex] m.

Its energy can be determined by;

E = hf

  = (hc) ÷ λ

where E is the energy, h is the Planck's constant, 6.626 × [tex]10^{-34}[/tex] Js, c is the speed of the light and f is its frequency.

Therefore,

E = (6.626 × [tex]10^{-34}[/tex]× 3.0 × [tex]10^{8}[/tex]) ÷ 1.0 × [tex]10^{-12}[/tex]

  = 1.9878 × [tex]10^{-25}[/tex] ÷ 1.0 × [tex]10^{-12}[/tex]

E = 1.9878 × [tex]10^{-13}[/tex] J

But, 1 eV = 1.6 × [tex]10^{-19}[/tex] J. So that;

E = [tex]\frac{1.9878*10^{-13} }{1.6*10^{-19} }[/tex]

  = 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.