Question

⁹⁸Mo(³He,d)⁹⁹Tc is the process used to produce the Tc isotope.

Calculate the Q-Value to three decimal places.

Calculate the maximum kinetic energy of the deuterium populating the ground state of Tc

Calculate the maximum kinetic energy of the deuterium populating the 0.18 MeV excited level of Tc

Calculate the maximum kinetic energy of the deuterium populating the 0.51 MeV excited level of Tc t

Answer 1

Please share the atomic weights of 98Mo , 99Tc , He, d.?

Q - Value for any reaction = (M initial - M final)c^2 = (M initial - M final) x 931.49 Mev

Note : 931.49 is the conversion factor of amu to c^2

98Mo(3He,d)99Tc is the process used to produce the Tc isotope.

(Lets calculate mass of LHS of this reaction - Mass of RHS of this reaction). Please note the below atomic weights :-

98Mo = 97.905 amu 99Tc = 98.906 amu

= He = 4.002603 amu    d = 2H = 2.014102 amu

Mass initial = 97.905 + 3 x 4.003 = 109.914 amu

Mass final = 98.906 + 2.014 = 100.92 amu

Q = (109.914 - 100.92) x 931.49 Mev [ Clearly there is an loss in mass which is released as energy. Hence a exothermic process.

= 8.378 Mev [ To three decimal places ] = Max KE of the deuterium populating the ground state of Tc

2.) At 0.18 Mev excited level of Tc

As Tc has energy 0.18 Mev more than Tc in the ground state, the K.E actually available is

Q = 8.378 - 0.18 Mev =8.198 Mev

3.) At 0.51 Mev excited level of Tc

As Tc has energy 0.51 Mev more than Tc in the ground state, the K.E actually available is

Q = 8.378 - 0.51 Mev = 8.368 Mev