Q: Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (\${t}_{1/2}\$) of the reaction.

For first order reaction,\$t = \cfrac{2.303}{k} \log{\cfrac{a}{a-x}}\$Given that:- \$x = 99.9 \; a\$ for \${t}_{99.9 \%}\$\$x = 0.5 \; a\$ for \${t}_{50 \%}\$\$\therefore \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.999a}}}{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.5a}}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\log{\cfrac{1}{0.001}}}{\log{\cfrac{1}{0.5}}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \log{0.001}}{\log{2}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \left( -3 \right)}{0.3010}\$\$\Rightarrow \; {t}_{99.9 \%} = 10 \times {t}_{50 \%}\$Hence proved.

Q: What is the wavelength of a photon emitted during a transition from \$n = 5\$ state to \$n = 2\$ state in the hydrogen atom?

\$\displaystyle \dfrac { 1}{ \lambda }=R [\dfrac { 1 }{n^2_1 } - \dfrac { 1}{ n^2_2 }] \$\$\displaystyle \dfrac { 1}{ \lambda }=109677.6 \ cm^{-1} \times [\dfrac { 1 }{2^2 } - \dfrac { 1}{ 5^2 }]\$\$\displaystyle \dfrac { 1}{ \lambda }=23032.3 \ cm^{-1} \$\$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm\$Convert unit from cm to nm1 cm \$\displaystyle =10^7 \ nm\$\$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm \times \dfrac {10^7 \ nm}{ 1 \ cm}\$\$\displaystyle \lambda = 434 \ nm\$

Q: A first order reaction takes \$40\$ min for \$30\$% decomposition. Calculate \$t_{1/2}\$.

Let \$a\$ M be the initial concentration. After \$40\$ minutes, the concentration is \$\displaystyle a - \dfrac {30a}{100} = 0.70a\$.\$\displaystyle k = \dfrac {2.303}{t}log\dfrac {[A]_o}{[A]} \$\$\displaystyle k =\dfrac {2.303}{40} log\dfrac {a}{0.70a}\$\$\displaystyle k = 8.92 \times 10^{-3} \: /min\$The half life period, \$\displaystyle t_{1/2} = \dfrac {0.693}{k} =\dfrac {0.693}{8.92 \times 10^{-3} \: /min} = 77.7\$ min.

Q: The half-life for radioactive decay of \$^{14}C\$ is \$5730\$ years. An archaeological artifact containing wood had only \$80\%\$ of the \$^{14}C\$ found in a living tree. Estimate the age of the sample.

Decay constant \$\displaystyle k = \frac {0.693}{t_{1/2}} = \frac {0.693}{5730 \: year} = 1.209 \times 10^{-4} \: /years\$The rate of counts is proportional to the number of C-14 atoms in the sample.\$\displaystyle N_0 = 100, N = 80\$The age of the sample \$\displaystyle t = \frac {2.303}{k} log\left( \frac {N_0}{N}\right)\$\$\displaystyle t = \frac {2.303}{1.209 \times 10^{-4}} \times log\left( \frac {100}{80}\right) = 1846 \: years\$

Q: \$75\%\$ of a first order reaction is completed in \$32\$ minutes. \$50\%\$ of the reaction will be completed in:

For a first-order reaction, we have:\$t_{75\%}=2t_{50\%}\$Given: \$t_{75\%}=32\$ minSubstituting the values in the above expression, we get-\$32\$ min \$=2\ t_{50\%}\$\$t_{50\%}=16\$ minsHence, \$50\%\$ of the reaction would have been completed in \$16\$ mins.

Q: For a first order reaction \$A \rightarrow B\$, the reaction rate at reactant concentration of 0.01 M is found to be \$2.0 \times10^{-5} mol L^{-1}s^{-1}\$. The half-life period of the reaction is:

For first order reaction,\$A \longrightarrow B\$Rate \$= k\times [A]\$Given, rate \$=2.0\times 10^{-5}mol \,L^{-1}s^{-1}\$[A] = Conc. of A = 0.01 M So, \$2.0\times 10^{-5}=k\times 0.01\$\$k=\frac{2.0\times 10^{-5}}{0.01}s^{-1}\$ \$ =2.0\times 10^{-3}s^{-1}\$For first order reaction,\$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.0\times 10^{-3}}\$\$=346.5=347s\$

Q: The value of 1 amu is equal to:

One an amu is \$\dfrac{1}{12}\$ of the mass of one carbon-12 atom. It is equal to \$ \dfrac{1}{N_A} \$\$= \dfrac{1}{6.022\times10^{23}} = 1.66 \times 10^{-24} g\$.

Q: Which of the following isotopes is used for treatment of cancer?

\$Co^{60}\$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Question 1

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (${t}_{1/2}$) of the reaction.

Accepted Answer

For first order reaction,$t = \cfrac{2.303}{k} \log{\cfrac{a}{a-x}}$Given that:- $x = 99.9 \; a$ for ${t}_{99.9 \%}$$x = 0.5 \; a$ for ${t}_{50 \%}$$\therefore \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.999a}}}{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.5a}}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\log{\cfrac{1}{0.001}}}{\log{\cfrac{1}{0.5}}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \log{0.001}}{\log{2}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \left( -3 \right)}{0.3010}$$\Rightarrow \; {t}_{99.9 \%} = 10 \times {t}_{50 \%}$Hence proved.

Question 2

What is the wavelength of a photon emitted during a transition from $n = 5$ state to $n = 2$ state in the hydrogen atom?

Accepted Answer

$\displaystyle  \dfrac {  1}{ \lambda }=R [\dfrac {  1 }{n^2_1  } - \dfrac {  1}{ n^2_2 }] $$\displaystyle  \dfrac {  1}{ \lambda }=109677.6 \ cm^{-1} \times  [\dfrac {  1 }{2^2  } - \dfrac {  1}{ 5^2 }]$$\displaystyle  \dfrac {  1}{ \lambda }=23032.3 \ cm^{-1} $$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm$Convert unit from cm to nm1 cm $\displaystyle =10^7 \ nm$$\displaystyle \lambda = 4.34 \times 10^{-5} \ cm \times  \dfrac {10^7 \ nm}{ 1 \ cm}$$\displaystyle \lambda = 434  \ nm$

Question 3

A first order reaction takes $40$ min for $30$% decomposition. Calculate $t_{1/2}$.

Accepted Answer

Let $a$ M be the initial concentration. After $40$ minutes, the concentration is $\displaystyle a - \dfrac {30a}{100} = 0.70a$.$\displaystyle k = \dfrac {2.303}{t}log\dfrac {[A]_o}{[A]} $$\displaystyle k =\dfrac {2.303}{40} log\dfrac {a}{0.70a}$$\displaystyle k = 8.92 \times 10^{-3} \: /min$The half life period, $\displaystyle t_{1/2} = \dfrac {0.693}{k} =\dfrac {0.693}{8.92 \times 10^{-3} \: /min} = 77.7$ min.

Question 4

The half-life for radioactive decay of $^{14}C$ is $5730$ years. An archaeological artifact containing wood had only $80\%$ of the $^{14}C$ found in a living tree. Estimate the age of the sample.

Accepted Answer

Decay constant $\displaystyle k = \frac {0.693}{t_{1/2}} = \frac {0.693}{5730 \: year} = 1.209 \times 10^{-4} \: /years$The rate of counts is proportional to the number of C-14 atoms in the sample.$\displaystyle N_0 = 100, N = 80$The age of the sample $\displaystyle t = \frac {2.303}{k} log\left( \frac {N_0}{N}\right)$$\displaystyle t = \frac {2.303}{1.209 \times 10^{-4}} \times log\left( \frac {100}{80}\right) = 1846 \: years$

Question 5

$75\%$ of a first order reaction is completed in $32$ minutes. $50\%$ of the reaction will be completed in:

Accepted Answer

For a first-order reaction, we have:$t_{75\%}=2t_{50\%}$Given: $t_{75\%}=32$ minSubstituting the values in the above expression, we get-$32$ min $=2\ t_{50\%}$$t_{50\%}=16$ minsHence, $50\%$ of the reaction would have been completed in $16$ mins.

Question 6

For a first order reaction $A \rightarrow B$, the reaction rate at reactant concentration of 0.01 M is found to be $2.0 \times10^{-5} mol L^{-1}s^{-1}$. The half-life period of the reaction is:

Accepted Answer

For first order reaction,$A \longrightarrow B$Rate $= k\times [A]$Given, rate $=2.0\times 10^{-5}mol \,L^{-1}s^{-1}$[A] = Conc. of A = 0.01 M So, $2.0\times 10^{-5}=k\times 0.01$$k=\frac{2.0\times 10^{-5}}{0.01}s^{-1}$ $ =2.0\times 10^{-3}s^{-1}$For first order reaction,$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.0\times 10^{-3}}$$=346.5=347s$

Question 7

The value of 1 amu is equal to:

Accepted Answer

One an amu is $\dfrac{1}{12}$ of the mass of one carbon-12 atom.  It is equal to $ \dfrac{1}{N_A} $$= \dfrac{1}{6.022\times10^{23}} = 1.66 \times 10^{-24} g$.

Question 8

Draw a graph showing the variation of potential energy between a pair of nucleons as a function of their separation. Indicate the regions in which the nuclear force is(i) Attractive(ii) Repulsive.Write two important conclusions which you can draw regarding the nature of the nuclear forces.

Accepted Answer

Conclusions:(i) The potential energy is minimum at a distance $r_0$ of about $0.8\,fm.$(ii) Nuclear force is attractive for distance larger than $r_0.$(iii) Nuclear force is repulsive if two are separated by distance less than $r_0.$(iv) Nuclear force decreases very rapidly at $r_0 / $equilibirium position.

Question 9

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 10

Which of the following isotopes is used for treatment of cancer?

Accepted Answer

$Co^{60}$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.