Number of atoms decayed in the first half-life = ½ (N_{0}) = ½ (500 × 10^{6}) = 250 × 10^{6} atoms.

Remaining amount = (500 – 250) × 10^{6} = 250 × 10^{6} = **250 million atoms.**

Similarly, in the second half-life, the atoms decayed are = ½ (250 × 10^{6}) = 125× 10^{6} atoms. The same amount will be left after 2^{nd} half-life, too.

And the number of atoms decayed in the 3^{rd} half-life = ½ (125 × 10^{6}) = 62.5 × 10^{6} atoms.

The amount of isotope remained after 3^{rd} half-life, 24 hours = (125 – 62.5) × 10^{6} = **62.5 × 10**^{6} atoms = 62 million atoms.

**Note:** An alternate method is, amount of the substance left after 3 half-lifes = (½)^{1/3 }of the total initial amount. Therefore, amount left after three half-lifes = (½)^{1/3} × 500 × 10^{6} = 0.125 × 500 × 10^{6} = 62.5 × 10^{6} = **62.5 million atoms.**

Pingback:Numerical Problem 3, Nuclear Physics – msa

Pingback:Numerical Problem 1, Nuclear Physics – msa

Pingback:Numerical Problems on Nuclear Physics – msa