Question
List- I shows different radioactive decay processes and List- II provides possible emitted particles. Match each entry in List- I with an appropriate entry from List- II , and choose the correct option. List- I List- II P 238 92 U → 234 91 P a 1 one α particle and one β + particle Q 214 82 P b → 210 82 P b 2 three β - particles and one α particle R 210 81 T l → 206 82 P b 3 two β - particles and one α particle S 228 91 P a → 224 88 R a 4 one α particle and one β - particle 5 one α particle and two β + particles
Step-by-step solution
In α decay mass number decreases by 4 unit and atomic number decreases by 2 unit. In β - decay mass number does not change but atomic number increases by 1 unit. In β + decay mass number does not change but atomic number decreases by 1 unit. So for P   238 92 U → 234 91 P a . Number of alpha particle = 238 - 234 4 = 1 . Due to alpha particle, atomic number should decrease by 2 but it has decreased by 1 only, therefore an additional β - decay is required, then net change in atomic number will be - 2 + 1 = - 1 . Therefore, P → 4 . Q   214 82 P b → 210 82 P b . Number of alpha particle = 214 - 210 4 = 1 . Due to alpha particle, atomic number should decrease by 2 , but there is no change in atomic number, which means two additional β - decay is required, then net change in atomic number will be - 2 + 2 = 0 . Therefore, Q → 3 . R   210 81 T l → 206 82 P b . Number of alpha particle = 210 - 206 4 = 1 . Due to alpha particle, atomic number should decrease by 2 , but atomic number has increased by 1 , which means three additional β - decay is required, then net change in atomic number will be - 2 + 3 = 1 . Therefore, R → 2 . S   228 91 P a → 224 88 R a . Number of alpha particle = 228 - 224 4 = 1 . Due to alpha particle, atomic number should decrease by 2 , but atomic number has decreased by 3 , which means one additional β + is required, then net change in atomic number will be - 2 - 1 = - 3 . Therefore, S → 1 .