Solveeit Logo

Question

Chemistry Question on Chemical Kinetics

The rate of radioactive disintegration at an instant for a radioactive same of half life 2.2×109s12.2 \times 10^9\,s^{-1} is 1010s110^{10} \,s^{-1}. The number of radioactive atoms in that sample at that instant is,

A

3.17×10203.17 \times 10^{20}

B

3.17×10173.17 \times 10^{17}

C

3.17×10183.17 \times 10^{18}

D

3.17×10193.17 \times 10^{19}

Answer

3.17×10193.17 \times 10^{19}

Explanation

Solution

T1/2=2.2×109s,R=1010s1,R=NλT_{1/2} = 2.2 \times 10^9 \,s, R = 10^{10} s^{-1} , R = N\lambda N=Rλ=R0.693T1/2N = \frac{R}{\lambda} = \frac{R}{0.693}T_{1/2} =1010×2.2×1090.693 = \frac{10^{10} \times 2.2 \times 10^9}{0.693} =3.17×1019 = 3.17\times 10^{19} atoms