Let Nβ be the number of β particles emitted by 1 gram of 24Na radioactive nuclei (half life=15 hrs) in 7.5 hours. Nβ is close to (Avogadro number=6.023×10²³/gmole)

If there are N radioactive nuclei at some time t, then the number dN which would decay in any given time interval dt would be proportional to N:
dN/dt = kN
[lnN]N₀ = -kt
N = N₀e⁻ᵏᵗ
given N = N₀/2 when t = t₁₂
K = ln2/t₁₂
finding the remaining molecules left after t = 7/12 hrs
N = N₀e⁻ᵏ(⁷/₁₂)
N = N₀√2
Number of particles emitted = N₀ - N₀/√2
N₀ - N₀/√2 = N₀(1 - 1/√2) = N₀(√2 -1)/√2
24Na × 6.023 × 10²³ = 7.34 × 10²¹, hence correct answer is option C