Agamma-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.

The energy and the wavelength of the photon are 1.546 MeV and .

Explanation:

Given that,

Kinetic energy = 261 KeV

Planck's constant

Speed of light

Mass of electron

Charge

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

Using formula of rest mass energy

Energy in eV

The total energy of photon

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

Using formula of wavelength

Put the value into the formula

Hence, The energy and the wavelength of the photon are 1.546 MeV and .

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