What is the maximum wavelength of the two photons produced when a positron annihilates an electron? The rest mass energy of each is 0.51 MeV.

(Answer: 2.44 ｘ 10⁻¹² m)



Given:

Rest mass energy of each photon = E =  0.51 eV = 0.51 ï½˜ 10⁶ ｘ 1.6 ｘ 10⁻¹⁹ J = 0.816  ｘ 10⁻¹³ J = 8.16  ｘ 10⁻¹⁴ J 

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Speed of light = 3 ｘ 10⁸ m s⁻¹


To Find:

Maximum Wavelength of the photon = λ =  ?


Solution:

Using the following formula for finding the wavelength of the incident photon as 

E = `\frac {hc}{λ}`

of

 Î» = `\frac {hc}{E}`

By putting values

 Î» = `\frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ 3 ｘ 10⁸ m s⁻¹ }{8.16 ｘ 10⁻¹⁴ J }`

 Î» = `\frac {19.878 ｘ 10⁻²⁶ m³ kg s⁻² }{8.16 ｘ 10⁻¹⁴ J }`

λ = 2.4360 ï½˜ 10⁻¹² m

or

λ = 2.44 pm -------------------Ans. 



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