The half−value period of a synthetic molecule is given to be 26.8 minutes. The mass of One curie of this molecule is (1 curie =3.71×1010dps and molecular mass of molecule =214 g)
A
3.71×1010g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.71×1010g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
8.61×1010g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
3.064×10−8g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D3.064×10−8g t12=26.8×60=1608s 1608=ln2τ 1608=ln21λ λ=ln21608 1curie=ln21608N0 3.7×1010=ln21608N0
∵N0=2146.022×1023×mg
Putting all the values we get mass of one curie of the molecule as 3.064×10−8g.