A highly energetic gamma-ray photon strikes a stationary deuteron ( H 2 1 ), causing it to undergo photodisintegration into a proton and a neutron. What is the maximum wavelength of the incident gamma-ray photon required to cause this disintegration? (Given: Atomic mass of H 2 1 = 2 . 01410 u , Atomic mass of H 1 1 = 1 . 00780 u , Mass of neutron = 1 . 00870 u , 1 u = 931 . 5 MeV / c 2 , and h c = 1242 MeV · fm )

Step-by-step solution

The photodisintegration reaction is: γ + H 2 1 → H 1 1 + n 1 0 First, calculate the mass defect ( Δ m ) for this process: Δ m = M ( H 1 1 ) + m n - M ( H 2 1 ) Δ m = ( 1 . 00780 + 1 . 00870 ) - 2 . 01410 Δ m = 2 . 01650 - 2 . 01410 = 0 . 00240   u The minimum energy (threshold energy) required to split the deuteron is: E = Δ m × 931 . 5   MeV / u E = 0 . 00240 × 931 . 5 = 2 . 2356   MeV This energy is provided by the incident gamma-ray photon. The