The half life of $U^{238}$ when 1 gram of it emits $1.23 \times 10^4$ alpha particle per second is: (taking 1 year $=3.155 \times 10^7$ second)

$4.519 \times 10^9$ years

The correct answer is Option (3) → $4.519 \times 10^9$ years

The activity A of Radioactive sample is given by -

$A=λN$

where,

A → Activity (particles per second) = $1.23×10^4$

λ → Decay constant

N → Number of radioactive present

and,

$λ=\frac{ln2}{T_{1/2}}$

and,

$N=\frac{1\,gram}{238\,grams/mol}×6.022×10^{23}atoms$

$=2.53×10^{21}atoms$

Now,

$λ=\frac{A}{N}=\frac{1.23×10^4}{2.53×10^{21}}\sec^{-1}$

$≃4.87×10^{-18}\sec^{-1}$

$∴T_{1/2}=\frac{ln2}{λ}=\frac{0.693}{4.87×10^{-18}}$

$≃1.42×10^{17}$ seconds $≃4.519×10^9$ years