JEE MAIN - Chemistry (2003 - No. 38)
The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of
it remaining undecayed after 18 hours would be
8.0 g
12.0 g
16.0 g
4.0 g
ব্যাখ্যা
$${t_{1/2}} = 3\,$$ hrs. $$T=18$$ hours
as $$\,\,\,\,T = n \times {t_{1/2}}$$
$$\therefore$$ $$\,\,\,\,n = {{18} \over 3} = 6$$
Initial mass $$\left( {{C_0}} \right) = 256\,g$$
$$\therefore$$ $$\,\,\,\,{C_n} = {{{C_0}} \over {{2^n}}} = {{256} \over {{{\left( 2 \right)}^6}}} = {{256} \over {64}} = 4g.$$
as $$\,\,\,\,T = n \times {t_{1/2}}$$
$$\therefore$$ $$\,\,\,\,n = {{18} \over 3} = 6$$
Initial mass $$\left( {{C_0}} \right) = 256\,g$$
$$\therefore$$ $$\,\,\,\,{C_n} = {{{C_0}} \over {{2^n}}} = {{256} \over {{{\left( 2 \right)}^6}}} = {{256} \over {64}} = 4g.$$