1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The mean lives of a radioactive substance are 1620 years and 405 years for α and β−emissions respectively. Then the time during which 14th of the sample will decay if it is decaying both α and β−emissions simultaneously will be:

A
92.4 years
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
449.2 years
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
106.0 years
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
172.0 years
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 92.4 yearsThis is an example of successive, emissions in parallel paths.λav=λα+λβ=11620+1405=51620yr−1As one fourth part decays, t=t,N=34NoThus, t=2.303λavlogNoN=2.303×16205log43=92.4yrs.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Neil Bohr Model
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program