Q.

In the case of Thorium (A = 232 and z = 90), we obtain an isotope of Lead (A = 208 and z = 82) after some radio active disintegrations. The number of β particles emitted are respectively

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Decrease of mass number = 232 - 208 = 24Number of α-particles emitted=244=6Due to emission of 6 particles, decrease in charge number is 12But actual decrease in charge number is 8.Clearly 12 - 8 = 4 β -particles are emitted
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon