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: 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: 99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

\$k = \dfrac{2.303}{t} log \dfrac{100}{100 - x}\$Substituting the values,\$k = \dfrac{2.303}{32} log \dfrac{100}{1}\$\$k = 0.144\space {min}^{-1}\$\$t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}\$\$t = 48\space min\$

Q: The charge on alpha particles is

Alpha particles have a charge of \$+2\$,Hence, charge on an alpha particle is twice the electronic charge \$ = 2 \times 1.6 \times 10^{-19} C\$\$q = 3.2 \times 10^{-19} C\$

Q: An \$\alpha \$-particle of energy 5 MeV is scattered through \${ 180 }^{ \circ }\$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Distance of closet approx\$:-\$\${r_0} = \dfrac{{ze\,\,2e}}{{4\pi { \in _0}E}}\$\$KE=PE\$\$\dfrac{1}{2}m{v^2} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{r}\$\$r = \dfrac{{9 \times {{10}^9} \times 2 \times 92 \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}\$\$r = 5.3 \times {10^{ - 14}}\,m \approx {10^{ - 12}}\,cm\$Hence,option \$(C)\$ is correct answer.

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

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 3

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 4

$75\%$ 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 5

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 6

Calculate the half life of a first order reaction from their rate constants given below:(a) $200 s^{-1}$; (b) $2\ min^{-1}$; (c) $4 year^{-1}$.

Accepted Answer

The relationship between the half life period and the rate constant is, $\displaystyle t_{1/2} = \frac {0.693}{k}$.(i) $\displaystyle t_{1/2} = \frac {0.693}{200} = 3.465 \times 10^{-3} \: s$(ii) $\displaystyle t_{1/2} = \frac {0.693}{2} = 0.3465 \: mins$(iii)  $\displaystyle t_{1/2} = \frac {0.693}{4} = 0.1732 \: years$

Question 7

99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

Accepted Answer

$k = \dfrac{2.303}{t} log \dfrac{100}{100 - x}$Substituting the values,$k = \dfrac{2.303}{32} log \dfrac{100}{1}$$k = 0.144\space {min}^{-1}$$t = \dfrac{2.303}{0.144} log \dfrac{100}{0.1}$$t = 48\space min$

Question 8

The charge on alpha particles is

Accepted Answer

Alpha particles have a charge of $+2$,Hence, charge on an alpha particle is twice the electronic charge $ = 2 \times  1.6 \times  10^{-19} C$$q = 3.2 \times  10^{-19} C$

Question 9

An $\alpha $-particle of energy 5 MeV is scattered through ${ 180 }^{ \circ  }$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Accepted Answer

Distance of closet approx$:-$${r_0} = \dfrac{{ze\,\,2e}}{{4\pi { \in _0}E}}$$KE=PE$$\dfrac{1}{2}m{v^2} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{q_1}{q_2}}}{r}$$r = \dfrac{{9 \times {{10}^9} \times 2 \times 92 \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{5 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}$$r = 5.3 \times {10^{ - 14}}\,m \approx {10^{ - 12}}\,cm$Hence,option $(C)$ is correct answer.

Question 10

The half life period of a first order reaction is $20$ minutes. The time required for the concentration of the reactant to change from $0.16 M$ to $0.02 M$ is:

Accepted Answer

The half life period of a first order reaction is 20 minutes. $\displaystyle t_{1/2} = 20$ minutes.The rate constant, $\displaystyle k =  \dfrac {0.693}{t_{1/2}}=\dfrac {0.693}{20 \: min}=0.03465 \: /min$The time required for the concentration of the reactant to change from 0.16M to 0.02Mis $\displaystyle t  =  \dfrac {2.303}{k}\log  \dfrac {a}{a-x}= \dfrac {2.303}{0.03465 \: /min}\log  \dfrac {0.16}{0.02} = 60 \: min$Hence, the correct option is $\text{B}$.

Revise with Concepts

Discovery of Radiation

Introduction to Radioactivity

Alpha, Beta and Gamma Emission

Differentiate Between Alpha, Beta and Gamma Radiations

Background Radiations

Quick Summary With Stories

Discovery of Radiation

Artificial radioactivity and application of radioactivity

Introduction to Radiation and its Sources

Radioactivity - Emission of Alpha, Beta, and Gamma Radiations

Differentiate Between Alpha, Beta and Gamma Radiations

Radiation Spectrum of Elements

Properties of Alpha, Beta and Gamma Radiations

Background Radiations

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.

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

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.

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

For a first order reaction $A \to B$ , the reaction rate at reactant concentration of 0.01 M is found to be $2.0 \times 1 0^{- 5} m o l L^{- 1} s^{- 1}$ . The half-life period of the reaction is:

Calculate the half life of a first order reaction from their rate constants given below:
(a) $200 s^{- 1}$ ; (b) $2 m i n^{- 1}$ ; (c) $4 y e a r^{- 1}$ .

99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

The charge on alpha particles is

An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

The half life period of a first order reaction is $20$ minutes. The time required for the concentration of the reactant to change from $0.16 M$ to $0.02 M$ is:

Revise with Concepts

Discovery of Radiation

Introduction to Radioactivity

Alpha, Beta and Gamma Emission

Differentiate Between Alpha, Beta and Gamma Radiations

Background Radiations

Quick Summary With Stories

Discovery of Radiation

Artificial radioactivity and application of radioactivity

Introduction to Radiation and its Sources

Radioactivity - Emission of Alpha, Beta, and Gamma Radiations

Differentiate Between Alpha, Beta and Gamma Radiations

Radiation Spectrum of Elements

Properties of Alpha, Beta and Gamma Radiations

Background Radiations

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (t1/2​) of the reaction.

A first order reaction takes 40 min for 30% decomposition. Calculate t1/2​.

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

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

For a first order reaction A→B, the reaction rate at reactant concentration of 0.01 M is found to be 2.0×10−5molL−1s−1. The half-life period of the reaction is:

Calculate the half life of a first order reaction from their rate constants given below: (a) 200s−1; (b) 2 min−1; (c) 4year−1.

99% of a first order reaction, was completed in 32 minute. When will 99.9% of the reaction complete?

The charge on alpha particles is

An α-particle of energy 5 MeV is scattered through 180∘ by a fixed uranium nucleus. The distance of the closest approach is of the order of

The half life period of a first order reaction is 20 minutes. The time required for the concentration of the reactant to change from 0.16M to 0.02M is:

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.

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

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.

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

For a first order reaction $A \to B$ , the reaction rate at reactant concentration of 0.01 M is found to be $2.0 \times 1 0^{- 5} m o l L^{- 1} s^{- 1}$ . The half-life period of the reaction is:

Calculate the half life of a first order reaction from their rate constants given below:
(a) $200 s^{- 1}$ ; (b) $2 m i n^{- 1}$ ; (c) $4 y e a r^{- 1}$ .

An $α$ -particle of energy 5 MeV is scattered through $180^{\circ}$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

The half life period of a first order reaction is $20$ minutes. The time required for the concentration of the reactant to change from $0.16 M$ to $0.02 M$ is: