Question

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.

Answer 1

Answer:

E = 1.24MeV

Explanation:

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

Its energy can be determined by;

E = hf

  = (hc) ÷ λ

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

Therefore,

E = (6.626 × $10^{-34}$× 3.0 × $10^{8}$) ÷ 1.0 × $10^{-12}$

  = 1.9878 × $10^{-25}$ ÷ 1.0 × $10^{-12}$

E = 1.9878 × $10^{-13}$ J

But, 1 eV = 1.6 × $10^{-19}$ J. So that;

E = $\frac{1.9878*10^{-13} }{1.6*10^{-19} }$

  = 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.