Question

6. The energy level diagram for 230 Tha+226Ra is shown below: 230Th 24% 0.068 - Co 76% 0 MeV 226 Ra Calculate the Q-value and the kinetic energy of the emitted alpha particle for alpha decay to the ground state of 226Ra. b) Calculate the Q-value and the kinetic energy of the emitted alpha particle for alpha decay to the excited state of Ra. If the excited state of 226Ra decays by gamma emission, use the energy level diagram to determine the wavelength of the emitted gamma ray.

Answer 1

Here is proof of energy transfer between alpha particle and daughter nucleus

Assume that initially parent atom is at rest

T denotes the energy of alpha and Daughter atom

for any alpha transition PD + x The (Q-value) is given by & = (mp-mo-ma) c² 230 Tha t Tha for ground state -0.068 Rar min = 230.033 u MRa - 226 u ma = 4.00 260 u a= [ 230.033 - 226 -4.0026] e² a = 0,0304 UIC2 Now, we know that lu=931.5 Mevlc? o = 28.308 Mev

ved Q = Txt To as momentum is also conserved explosion so after a Px = To - 0 I Pa = Vama Ta , Po= Vamoto put in m o Та – тр Тр ma - A la Ta ) 4 where A +4 A is mass of daugter nude

8 Now, this QL is distributed to a partide and dauter nuclei A = 226u KEx = 226 y Q -226x 28.308 230 230 KEL = 27.81 Mev 16 Now, our Ra is in excited state so Qx' = Qd - (energy stored matom) 28. 308-0.668 28.24 Mev KEL = 226 Qx' = 226 x 28.24 230 230 27.748 Mev (c. as Ra transit to ground state after realising the r rays so energy of & says is 0.068 Mev

Er = ho hy he Er we know hc=12423.75 EVA put in y = -12423.75 0.068 x 106 ev = 0.18 *ro-10 m kg = 1.8x10-um.