Nuclear Structure — Important Questions

19.9%

Let the lighter isotope abundance be P1 and the other be P2 = 100 - P1. Using average atomic mass, 10.811 = (P1 x 10.01294 + (100 - P1) x 11.00931)/100. Solving gives P1 ≈ 19.9%.

20.18 u

Average atomic mass = (90.51 x 19.99 + 0.27 x 20.99 + 9.22 x 21.99)/100. This gives 20.18 u.

A neutron is emitted, The reaction is 9Be + 4He -> 10C + neutron + Q

The neutron discovery reaction is 9Be + 4He -> 10C + neutron + Q. It directly shows that a neutron is the emitted particle.

1H

Balance mass number: 14 + 4 = 18, so the unknown particle must have mass number 1 because 17 + 1 = 18. Balance atomic number: 7 + 2 = 9, so the unknown particle must have atomic number 1. That is a proton, 1H.