JEE Main 16 March 2021 Shift 2 Solved Paper

Given,
Half-life of Au198,T1/2=2.7 days
Amount of Au198,m=1.5mg=1.5×10−3g
Atomic weight of Au198,A=198gmol−1
We know that activity, A0=λN0
where, λ= decay constant =‌

ln‌2

T1/2

N0= initial number of atoms =‌

m

A

NA
⇒=A0λ⋅‌

m

A

NA=‌

ln‌2

T1∕2

×‌

1.5×10−3

198

×6.023×1023
⇒A0=‌

ln‌2×1.5×10−3×6×1023

2.7×3600×24×198×3.7×1010

Ci
[∵1 curie (Ci)=3.7×1010Bq]
⇒A0=3.65×102Ci=365Ci
The answer is close to 357Ci.