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Differentiate Between Alpha, Beta and Gamma Radiations

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: 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}\$.

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\$

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.

Q: 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:

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}\$.

Physics

definition

Differentiate between alpha, beta and gamma radiations

Property	$α$ -Particle	$β$ -Particle	$γ$ -Particle
Nature	Stream of positively charged particle i.e. helium nuclei	Stream of negatively charged particles, i.e. energetic electrons	Highly energetic electromagnetic radiation
Speed	Nearly $1 0^{7} m s^{- 1}$	About 90% of the speed of light or $2.7 \times 1 0^{8} m s^{- 1}$	$3 \times 1 0^{8} m s^{- 1}$ in vacuum
Penetrating Power	Small (3 to 8cm in air)	Large (Up to few metre in air)	Very Large (Up to a few hundred metre in air)

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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:

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definition

Differentiate between alpha, beta and gamma radiations

REVISE WITH CONCEPTS

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Differentiate Between Alpha, Beta and Gamma 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: