State the law of radioactive decay. Hence derive the expression \$N={ N }_{ 0 }{ e }^{ -\lambda t }\$ where symbols have their usual meanings.

Law of radioactive decay : At any instant, the rate of radioactive disintegration is directly proportional to the number of nuclei of the radioactive element present at that instant.We know by radioactive decay law\$\dfrac { dN }{ dt } \propto N\$or \$\dfrac { dN }{ dt } =-\lambda N\$where \$\lambda\$ is the constant of proportionality called radioactive decay constant. Let \$ { N }_{ 0 }\$ be the number of nuclei present at time \$t=0\$, and \$N\$ the number of nuclei present at time \$t\$.\$\int _{ { N }_{ 0 } }^{ N }{ \dfrac { dN }{ N } } =-\int _{ 0 }^{ t }{ \lambda dt } =-\lambda \int _{ 0 }^{ t }{ dt }\$\$ \log _{ e }{ N } -\log _{ e }{ { N }_{ 0 } } =-\lambda t\$\$\log _{ e }{ \left( \dfrac { N }{ { N }_{ 0 } } \right) } =-\lambda t\$\$\dfrac { N }{ { N }_{ 0 } } ={ e }^{ -\lambda t }\$\$N={ N }_{ 0 }{ e }^{ -\lambda t }\$

The half life of radioactive radon is 3.8 days. The time at the end of which \$\displaystyle\frac{1}{20}\$th of the Radon sample will remain undecayed is (given \$\log _{10}e=0.4343\$)

\$\displaystyle t_{1/2}=3.8\$ day\$\displaystyle \therefore \lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{3.8}=0.182\$If the initial number of atom is \$\displaystyle a=A_{0}\$ then after time t the number of atoms is \$a/20=A\$.\$\displaystyle t=\frac{2.303}{\lambda}\log \frac{A_{0}}{A}=\frac{2.303}{0.182}\log \frac{a}{a/20}\$\$\displaystyle =\frac{2.303}{0.182}\log 20=16.46\$

Radioactive material '\$A\$' has decay constant '\$8\lambda\$' and material '\$B\$' has decay constant '\$\lambda\$'. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be \${e}\$?

\$N_A=N_{A_o}e^{(-8\lambda)t}\$\$N_B = N_{B_o}e^{-\lambda t}\$\$N_{A_o} = N_{B_0}\$\$\Rightarrow \dfrac{N_B}{N_A} = e\$\$\Rightarrow \dfrac{e^{-\lambda t}}{e^{-8\lambda t}} = e^{1}\$\$\Rightarrow e^{7\lambda t } = e^{1}\$\$7\lambda t = 1\$\$t = \dfrac{1}{7\lambda}\$

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Plot Number of Neutrons vs Protons for Radioactive Decay

Q: Time required to decompose \$S{O_2}C{L_2}\$ to half of itsinitial amount is 60 minutes . If the decompose is a first order reaction,calculate the rate constant of thereaction.

Given \$\therefore t_{1/2}= 60\, min= 60\times 60= 3600 \, sec\$\$\because \$ we know that 1 min = 60 secHalf life for the \$1^{st}\$ order reaction,\$t_{1/2}=\dfrac{0.693}{k}\$\$3600= \dfrac{0.693}{k}\$\$k = \dfrac{0.693}{3600}= 0.0001925= 1.925\times 10^{-4}\, s^{-1}\$

Q: The half-life of a radioactive substance is \$30\$ min. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is:

Ratio of remaining nuclei (N) to initial nuclei (No) having half life period of \$T=30\$ min at time \$t\$ is given by :-\$\dfrac{N}{N_0}=\left(\dfrac{1}{2}\right)^{t/T}\$\$\Rightarrow t=-T \log_2\left(\dfrac{N}{N_0}\right)\$(i) when \$40\%\$ decay completes in time \$t_1\$(i.e \$\dfrac{N}{N_0}0.6)\$ [\$40\%\$ decay means ,\$60\%\$ remaining ]\$\Rightarrow t_1=-T \log_2(0.6)\$(ii) when \$85\%\$ decay completes in time \$t_2\left(\dfrac{N}{N_0}=0.15\right)\$\$t_2=-T\log_2(0.15)\$So time taken ,\$\Delta t =t_2-t_1=T[\log_2(0.6)-\log_2(0.15)]\$\$=T \log_2\left(\dfrac{0.6}{0.15}\right)\$\$=30\times \log_2(4)\$\$\Delta t=60 min\$

Q: The half-life of radium is 1600 years. The fraction of a sample of radium that would remain after 6400 years :

As radioactive decay is first-order kinetichalf-life given is \$1600\$ years in \$6400\$ no. of half-life isn\$= \dfrac{6400}{1600}=4\$fraction of element remains after n half life is given by \$=\dfrac{1}{2^n}\$Hence fraction of radium remains after \$4\$ half life\$=\dfrac{1}{2^4}=\dfrac{1}{16}\$.Option D.

Q: A sample of a radioactive element has a mass of \$10 g\$ at an instant \$t=0\$. The approximate mass of this element in the sample after two mean lives is

Mean Life \$=\cfrac{1}{\lambda}\$\$N_o = 10 gms\$First order decay:\$N = N_oe^{- \lambda t}\$At \$t = \cfrac{2}{\lambda}\$\$N = N_oe^{-2} = 1.35\ gms\$

Q: For a radioactive material, half-life is \$10\$ minutes. If initially there are \$600\$ number of nuclei, the time taken (in minutes) for the disintegration of \$450\$ nuclei is?

Half life \$ = 10min = \dfrac {ln2}{\lambda}\$Nuclei disintegrate = 450Nuclei left = 600-450 =150\$N=N_{o}e^{-\lambda}{t}\$\$150 = 600 e^{\dfrac{-ln2}{10} t}\$so, t = 20 minutes

Q: The half-life of \$_{ 92 }^{ 238 }{ U }\$ undergoing \$\alpha-\$decay is \$4.5\ \times 10^{9}\$ years. What is true activity of the \$1g\$ sample of \$ _{ 92 }^{ 238 }{ U }\$? (Avogadro number\$=6.022\ \times 10^{23}\$)

Given that, Half life \${{T}_{\frac{1}{2}}}=4.5\times {{10}^{9}}=4.5\times {{10}^{9}}\times 3.156\times {{10}^{7}}s\$ Avogadro number \$N=6.023\times {{10}^{23}}\$ Now, the number of atoms in 1g of uranium \$N=\dfrac{6.023\times {{10}^{23}}}{238}\$Now, the activity of the sample \$ =\lambda N \$ \$ =\dfrac{0.693}{{{T}_{\frac{1}{2}}}}\times N \$ \$ =\dfrac{0.693}{4.5\times {{10}^{9}}\times 3.15\times {{10}^{7}}}\times \dfrac{6.023\times {{10}^{23}}}{238} \$ \$ =1.237\times {{10}^{4}}\,d/s \$ Hence, the activity of the sample is \$1.237\times {{10}^{4}}\,d/s\$

Q: A radioactive isotope has a half-life of \$T\$ years. How long will it take the activity to reduce to (a) \$3.125\%\$ (b) \$1\%\$ of its original value?

(a) Let the half-life be \$T\$ years.original amount of isotope is \$N_o\$after decay, the remaining amount is \$N\$now as per given values, \$N/N_o=3.125\%=1/32\$also, \$N/N_o=e^{-\lambda t}\$where \$\lambda\$ is decay constantalso \$\lambda=0.693/T\$so, \$t=3.4657/\lambda=5T\$ years.(b) Let the half-life be \$T\$ years.original amount of isotope is \$N_o\$after decay, the remaining amount is \$N\$now as per given values, \$N/N_o=1\%=1/100\$also, \$N/N_o=e^{-\lambda t}\$where \$\lambda\$ is decay constantalso \$\lambda=0.693/T\$so, \$t=4.6052/\lambda=6.645T\$ years.

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