Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

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

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 2.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon