Skip to main content
Back to NUCLEI

NEET UG PHYSICS MCQ: NUCLEI

1 Medium
1 of 1
Application +4 | -1
For thorium A=232A = 232,Z=90Z = 90 at the end of some radioactive disintegration we obtain an isotope of lead with A=208A = 208 and Z=82Z = 82, then the number of emitted α  and  β\mathrm{\alpha }\,\,\mathrm{and}\,\,\mathrm{\beta } particles are
View answer and explanation
Correct option: C
α=6,β=4\mathrm{\alpha } = 6,\mathrm{\beta } = 4
Option C is correct because alpha decay reduces the mass number by 4 and the atomic number by 2, while beta decay leaves the mass number unchanged and increases the atomic number by 1. By setting up equations based on the changes in mass and atomic numbers, we can solve for the number of alpha and beta particles emitted.

More Questions from NUCLEI

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.