Question

In the uranium radioactive series, the initial nucleus is $_{92}U^{238}$ and that the final nucleus is $_{82}Pb^{206}$. When uranium nucleus decays to lead, the number of α particles and β particles emitted are(a) 8α, 6β(b) 6α, 7β(c) 6α, 8β(d) 4α, 3β

Answer 1

Heya mate!!

Option (A) is correct.

Answer 2

Answer:

8

Explanation:

Initial nucleus _{92}U^{238} (Given)

Final nucleus -  _{82}Pb^{206} (Given)

U238→Pb206

An alpha particle has two neutrons and two protons(A helium nucleus)

Thus the combined mass will be 4

Decay -

= 206  =  238−  (4x)

Subtracting 238 from both the sides

= 238-4x-238 = 206-238

= -4x = -32

= x = 8

This means that this process undergoes 8 alpha or  α  decays which means 8 nuclei have been emitted. Thus the alpha decays will be -