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: A radioactive nucleus of mass \$M\$ emits a photon of frequency \$v\$ and the nucleus recoils. The recoil energy will be

Energy of emitted photon is \$E=h\nu\$ For a photon its momentum is given as \$\dfrac{Energy}{velocity \ of \ light}\$ which is nothing but \$\dfrac{E}{c}\$ or \$\dfrac{h\nu}{c}\$ According to momentum conservation value of the momentum of recoiled nucleus should also be \$\dfrac{h\nu}{c}\$ but in opposite direction to make net momentum zero before and after emission.Now recoil energy \$E_2\$ = \$\dfrac { p^2}{2M}\$ where \$p\$ is recoil momentum putting \$p\$ \$=\$ \$\dfrac{h\nu}{c}\$ we get recoil energy as \$\dfrac{h^{2} \nu^{2}}{2Mc^2}\$ i.e., option C

Q: In a first order reaction the concentration of reactant decreases from \$800 mol/{dm}^{3}\$ to \$50 mol/{dm}^{3}\$ in \$2\times {10}^{2}s\$. The rate constant of reaction in \${s}^{-1}\$ is:

\$\Rightarrow\$ \$k=\cfrac{2.303}{t}\log _{ 10 }{ \cfrac { a }{ a-x } } \$\$\because\$ time \$t=2\times {10}^{2}s\$\$a=800mol\$ \$ {dm}^{-3}\$\$a-x=50mol\$ \$ {dm}^{-3}\$\$\Rightarrow\$ \$k=\cfrac { 2.303 }{ 2\times { 10 }^{ 2 } } \log _{ 10 }{ \cfrac { 800 }{ 50 } } \$\$\Rightarrow\$ \$k=\cfrac { 2.303 }{ 2\times { 10 }^{ 2 } } \log _{ 10 }{ 16 } \$\$\therefore\$ \$k=1.38\times {10}^{-2}{s}^{-1}\$

Q: A radioactive sample at any instant has its disintegration rate \$5000\$ disintegration per minute. After \$5\$ minutes, the rate is \$1250\$ disintegrations per minute. Then, the decay constant, per minute, is

Let decay constant per minute\$=\lambda\$Therefore, ATQ, initially disintegration rate,\$R={ N }_{ \circ }\lambda =5000\rightarrow (1)\\ \$Also, in 2nd case, disintegration rate after 5 min is,\$1250=N\lambda\rightarrow (2)\$Dividing equation (1) by (2), we get\$\frac { { N }_{ \circ }\lambda }{ N\lambda } =\frac { 6000 }{ 1250 } \\ \frac { { N }_{ \circ } }{ N } =4\\ \$And as we know, the number of decay, N is given by,\$N={ N }_{ \circ }{ e }^{ -\lambda t }\quad (t=5\quad min)\\ \frac { 1 }{ 4 } ={ e }^{ -5\lambda }\\ -ln(4)=-5\lambda \\ \lambda =0.4ln2\$

Q: The activity of a radioactive sample is measured as 9750 count/min at t = 0 and 975 count/min at t = 5 min.The decay constant is nearly

Initial activity of the sample \$R_o = 9750\$ count/minActivity after \$t=5\$ min, \$R = 975\$ count/minUsing First order decay, \$R = R_o e^{-\lambda t}\$\$\therefore\$ \$975 = 9750 e^{-5\lambda }\$Or \$ - \ln 10 = -5\lambda\$Or \$ 2.303 = 5\lambda\$\$\implies\$ \$\lambda = 0.461\$ \$min^{-1}\$

Q: Activity of a radioactive element is \$10^{3}\$ dps. Its half life is \$1\ sec\$. After \$3\ sec\$ its activity will be (\$dps=decay \ per \ second\$)

\$A = Ao_e^{-\lambda T}\$ \$\lambda = \dfrac{0.693}{1}\$\$= 10^3 \cdot e^{-2.079}\$\$= \dfrac{10^3}{e^{2.079}} = \dfrac{1000}{8} = 125\$ dps.

Question 1

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

Accepted Answer

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 }$

Question 2

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?

Accepted Answer

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

Question 3

A radioactive nucleus of mass $M$ emits a photon of frequency $v$ and the nucleus recoils. The recoil energy will be

Accepted Answer

Energy of emitted photon is $E=h\nu$ For a photon its momentum is given as $\dfrac{Energy}{velocity \ of \ light}$ which is nothing but $\dfrac{E}{c}$ or $\dfrac{h\nu}{c}$ According to momentum conservation value of the momentum of recoiled nucleus should also be $\dfrac{h\nu}{c}$  but in opposite direction to make net momentum zero before and after emission.Now recoil energy $E_2$ = $\dfrac { p^2}{2M}$ where $p$ is recoil momentum putting $p$ $=$ $\dfrac{h\nu}{c}$ we get recoil energy as $\dfrac{h^{2} \nu^{2}}{2Mc^2}$ i.e., option C

Question 4

In a first order reaction the concentration of reactant decreases from $800 mol/{dm}^{3}$ to $50 mol/{dm}^{3}$ in $2\times {10}^{2}s$. The rate constant of reaction in ${s}^{-1}$ is:

Accepted Answer

$\Rightarrow$ $k=\cfrac{2.303}{t}\log _{ 10 }{ \cfrac { a }{ a-x }  } $$\because$ time $t=2\times {10}^{2}s$$a=800mol$ $ {dm}^{-3}$$a-x=50mol$ $ {dm}^{-3}$$\Rightarrow$ $k=\cfrac { 2.303 }{ 2\times { 10 }^{ 2 } } \log _{ 10 }{ \cfrac { 800 }{ 50 }  } $$\Rightarrow$ $k=\cfrac { 2.303 }{ 2\times { 10 }^{ 2 } } \log _{ 10 }{ 16 } $$\therefore$ $k=1.38\times {10}^{-2}{s}^{-1}$

Question 5

A radioactive sample at any instant has its disintegration rate $5000$ disintegration per minute. After $5$ minutes, the rate is $1250$ disintegrations per minute. Then, the decay constant, per minute, is

Accepted Answer

Let decay constant per minute$=\lambda$Therefore, ATQ, initially disintegration rate,$R={ N }_{ \circ  }\lambda =5000\rightarrow (1)\ $Also, in 2nd case, disintegration rate after 5 min is,$1250=N\lambda\rightarrow (2)$Dividing equation (1) by (2), we get$\frac { { N }_{ \circ  }\lambda  }{ N\lambda  } =\frac { 6000 }{ 1250 } \ \frac { { N }_{ \circ  } }{ N } =4\ $And as we know, the number of decay, N is given by,$N={ N }_{ \circ  }{ e }^{ -\lambda t }\quad (t=5\quad min)\ \frac { 1 }{ 4 } ={ e }^{ -5\lambda  }\ -ln(4)=-5\lambda \ \lambda =0.4ln2$

Question 6

Two radioactive sources A and B initially contain equal number of radioactive atoms. Source A has a half-life of 1 hour and source B has a half-life of 2 hours. At the end of 2 hours, the ratio of the rate of disintegration of A to that of B is :

Accepted Answer

Two radioactive nuclei X and Y initially contain equal number of atoms. The half life is 1 hour and 2 hours respectively. Calculate the radio of their rates of disintegration after two hours.Given, nucleii X and Y contain equal number of atoms.Half-life of X, $T_{1}= 1 hr$Half-life of Y, $T_{2}=2 hr$Radio of radioactive sample of X left after 2 hours,$N_{1}=(\frac{1}{2})^{2/1}N_{0}=\frac{1}{4}N_{0}$Radio of radioactive sample of Y left after 2 hours,$N_{2}=(\frac{1}{2})^{2/2}N_{0}=\frac{1}{2}N_{0}$Now, decay rate is given by,$R=\lambda N=\frac{0.693 N}{T_{1/2}}$$\Rightarrow R\propto \frac{N}{T_{1/2}}$$\frac{R_{1}}{R_{2}}=\frac{(T_{1/2})_{2}}{(T_{1/2})_{1}}\times \frac{N_{1}}{N_{2}}$i.e., $=\frac{2}{1}\times \frac{\frac{N_{0}}{4}}{\frac{N_{0}}{2}}$$=1$

Question 7

The relation between half life ($T$) and decay constant ($\lambda$) is

Accepted Answer

As per radiaoactive decay law, the radioactive decays per unit time are directly proportional to the &#10;total number of nuclei of radioactive compounds in the sample. f the number of nuclei in a sample is N and the number of radioactive decays per unit time dt is dN then,\begin{equation}\frac{dN}{dt}=-\lambda N\end{equation}Integrating both limits\begin{equation}\int_{N_0}^N\frac{dN}{N}=&#8211;\lambda ~\int \limits^t_{t_0}dt\end{equation}Taking initial time $t_0$=0 we have\begin{equation}ln\frac{N}{N_0}=-\lambda t\end{equation}By definition of half life ( time period in which  the number of nuclei, N, decayed by half value) i.e.  $N_t$=$\dfrac{N_0}{2}$$\therefore$$ln2=\lambda t=\log_e 2$

Question 8

The activity of a radioactive sample is measured as 9750 count/min at t = 0 and 975 count/min at t = 5 min.The decay constant is nearly

Accepted Answer

Initial activity of the sample  $R_o = 9750$ count/minActivity after $t=5$ min,  $R = 975$  count/minUsing First order decay,  $R = R_o e^{-\lambda t}$$\therefore$   $975 = 9750 e^{-5\lambda }$Or  $ - \ln 10 = -5\lambda$Or  $ 2.303 = 5\lambda$$\implies$  $\lambda = 0.461$ $min^{-1}$

Question 9

Define the term: Half-life period and decay constant of a radioactive sample.Derive the relation between these terms.

Accepted Answer

Half-life period: The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.Decay constant: The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to $\dfrac{1}{e}$ times of its initial time.Relation between Half-life and Decay constant: The radioactive decay equation is                $N = N_{0}e^{-\lambda t}$         ........(i)when $t = T, N = \dfrac{N_0}{2}$$\therefore \dfrac{N_0}{2} = N_0 e^{-\lambda T}$or   $e^{-\lambda T} = \dfrac{1}{2}$      .....(ii)Taking log of both sides               $-\lambda T log_{e} = log_{e} 1 - log_{e} 2$or  $\lambda T = log_{e} 2$$\therefore T = \dfrac{log_e 2}{\lambda} = \dfrac{2.3026 \,log_{10}2}{\lambda} = \dfrac{2.3026 \times 0.3010}{\lambda}$     ........(iii)or  $T = \dfrac{0.6931}{\lambda}$

Question 10

Activity of a radioactive element is $10^{3}$ dps. Its half life is $1\ sec$. After $3\ sec$ its activity will be ($dps=decay \ per \ second$)

Accepted Answer

$A = Ao_e^{-\lambda T}$           $\lambda = \dfrac{0.693}{1}$$= 10^3 \cdot e^{-2.079}$$= \dfrac{10^3}{e^{2.079}} = \dfrac{1000}{8} = 125$ dps.

Problems on Radioactivity - II.

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