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← IIT-JEE 2009 · Paper 2
ChemistryInteger+4 / -1Question 15 of 57

IIT-JEE 2009 · Paper 2, Chemistry question 15

The total number of α\alpha and β\beta particles emitted in the nuclear reaction X92238X2922238U→X82214X2822214Pb\ce{^{238}_{92}U -> ^{214}_{82}Pb} is

Answer and worked solution

Official answer: 8

α particles. The mass number falls by 238−214=24238-214=24: six α particles, which lower Z by 12 to 80.

β particles. Z must rise from 80 to 82: two β⁻ particles.

Total 6+2=86+2=8.

Answer: 8