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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Half Life

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.

Physics

definition

Half life of a radioactive substance

The half-life

T_{1 / 2}

of a radio nuclide is the time at which both N and R have been reduced to one-half their initial values.
Putting

R = (1 / 2) R_{0}

and

t = T_{1 / 2}

in the law of radioactive decay, we get:

T_{1 / 2} = \frac{l n 2}{λ} = \frac{0 . 6 9 3}{λ}

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Important Questions

Time required to decompose $S O_{2} C L_{2}$ to half of its
initial amount is 60 minutes . If the decompose is a first order reaction,
calculate the rate constant of the
reaction.

Medium

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

Medium

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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:

Medium

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

Hard

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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

Medium

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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?

Medium

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The half life of radioactive radon is 3.8 days. The time at the end of which $\frac{1}{2 0}$ th of the Radon sample will remain undecayed is (given $lo g_{10} e = 0.4343$ )

Medium

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The half-life of $_{92}^{238} U$ undergoing $α -$ decay is $4.5 \times 1 0^{9}$ years. What is true activity of the $1 g$ sample of $_{92}^{238} U$ ? (Avogadro number $= 6.022 \times 1 0^{23}$ )

Medium

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

Medium

NEET

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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?

Medium

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definition

Half life of a radioactive substance

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Law of Radioactive Decay

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Half Life and Mean Life of Radioactive Substances

Time required to decompose SO2​CL2​ to half of its initial amount is 60 minutes . If the decompose is a first order reaction, calculate the rate constant of the reaction.

State the law of radioactive decay. Hence derive the expression N=N0​e−λt where symbols have their usual meanings.

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:

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

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

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?

The half life of radioactive radon is 3.8 days. The time at the end of which 201​th of the Radon sample will remain undecayed is (given log10​e=0.4343)

The half-life of 92238​U undergoing α−decay is 4.5 ×109 years. What is true activity of the 1g sample of 92238​U? (Avogadro number=6.022 ×1023)

Radioactive material 'A' has decay constant '8λ' and material 'B' has decay constant 'λ'. 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?

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?

Time required to decompose $S O_{2} C L_{2}$ to half of its
initial amount is 60 minutes . If the decompose is a first order reaction,
calculate the rate constant of the
reaction.

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

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:

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

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?

The half life of radioactive radon is 3.8 days. The time at the end of which $\frac{1}{2 0}$ th of the Radon sample will remain undecayed is (given $lo g_{10} e = 0.4343$ )

The half-life of $_{92}^{238} U$ undergoing $α -$ decay is $4.5 \times 1 0^{9}$ years. What is true activity of the $1 g$ sample of $_{92}^{238} U$ ? (Avogadro number $= 6.022 \times 1 0^{23}$ )

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

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?