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

E = 1.24MeV

Step-by-step 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 =
`(1.9878*10^(-13) )/(1.6*10^(-19) )`

= 1242375 eV

∴ E = 1.24MeV

The energy of the photon is 1.24MeV.