COMEDK2013ChemistryChemical Kinetics
The number of and -particles emitted during the transformation of ₉₀ Th ²³² to ₈₂ ~Pb ²⁰⁸ are respectively
Options
- A4,2
- B2,2
- C8,6
- D6,4
Correct answer
D. 6,4
Step-by-step solution
₉₀²³² Th ₈₂²⁰⁸ ~Pb +n +m Number of -particles, n= 232-208 4 =6 Now, for charge balanced aligned 90 &=82+2 n-m m &=4 aligned Number of -particles, m=4