Question

A gamma-ray photon produces an electron and a positron, each with a kinetic energy of 261 keV . h=6.626×10−34J⋅s, c=2.998×108m/s, me=9.109×10−31kg, e=1.602×10−19C.

A) Determine the energy of the photon.B) Determine the wavelength of the photon.

Answer 1

The energy and the wavelength of the photon are 1.546 MeV and
`8.036*10^(-13)\ m`.

Step-by-step explanation:

Given that,

Kinetic energy = 261 KeV

Planck's constant
`h = 6.626*10^(−34)\ J.s`

Speed of light
`c=2.998*10^(8)\ m/s`

Mass of electron
`m_(e)=9.109*10^(-31)\ kg`

Charge
`q=1.602*10^(-19)\ C`

(A). We need to calculate the energy of the photon

Using formula of rest mass energy

`E=m_(0)c^2`

`E=9.109*10^(-31)*(3*10^(8))^2`

`E=8.198*10^(-14)\ J`

Energy in eV

`E=(8.198*10^(-14))/(1.6*10^(-19))`

`E=512375\ eV`

`E=0.512\ MeV`

The total energy of photon

`TE=2(E+K.E)`

`TE=2(0.512+0.261)`

`TE=1.546\ MeV`

(B). We need to calculate the wavelength of the photon

Using formula of wavelength

`\lambda=(hc)/(E)`

Put the value into the formula

`\lambda=(6.626*10^(−34)*3*10^(8))/(1.546*10^(6)*1.6*10^(-19))`

`\lambda=8.036*10^(-13)\ m`

Hence, The energy and the wavelength of the photon are 1.546 MeV and
`8.036*10^(-13)\ m`.