147644
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. Calculate the minimum time after which it would be possible to work safely with this source.
147646
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
1 $\frac{1}{\mathrm{e}^{2}}$
2 $\frac{1}{\mathrm{e}}$
3 $\mathrm{e}$
4 $\mathrm{e}^{2}$
Explanation:
C Let, the initial number of atom $=\mathrm{N}_{0}$ The number of atom at an instant of time $=\mathrm{N}$ So, mean life $(\tau)=\frac{1}{\lambda}$ Given, Number of atom at an instant time $(\mathrm{t})=$ mean life $(\tau)$ According to radioactive law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \quad \ldots . . \text { (ii) } \quad(\therefore \mathrm{t}=1 / \lambda)$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\left(\lambda \times \frac{1}{\lambda}\right)}$ From equation (i) and equation (ii) $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$
Manipal UGET-2009
NUCLEAR PHYSICS
147647
The half-life of $\mathrm{At}^{215}$ is $100 \mu$ s. If a sample contains $215 \mathrm{mg}$ of $\mathrm{At}^{215}$, the activity of the sample initially is
1 $10^{2} \mathrm{~Bq}$
2 $3 \times 10^{10} \mathrm{~Bq}$
3 $4.17 \times 10^{24} \mathrm{~Bq}$
4 $1.16 \times 10^{5} \mathrm{~Bq}$
Explanation:
C Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=100 \mu \mathrm{s}=100 \times 10^{-6} \mathrm{~s}=10^{-4} \mathrm{~s}$ $\lambda=\frac{0.6931}{t_{1 / 2}}$ $\lambda=\frac{0.6931}{10^{-4}}=0.6931 \times 10^{4} \mathrm{~s}^{-1}$ Number of atoms in $215 \mathrm{mg}$ of sample $(\mathrm{N})-$ $=\frac{6.023 \times 10^{23} \times 215 \times 10^{-3}}{215}$ $\mathrm{~N}=6.023 \times 10^{20}$ Activity, $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=0.6931 \times 10^{4} \times 6.023 \times 10^{20}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=4.17 \times 10^{24} \mathrm{~Bq}$
Manipal UGET-2010
NUCLEAR PHYSICS
147648
Half-life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used ?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 4.18 day
4 1.2 day
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=3.20 \mathrm{~h}$ Percentage of substance is used $=75 \%$ So, Percentage of substance left $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=1-\frac{75}{100}$ $=25 \%$ $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=2$ $\mathrm{t}=2 \times \mathrm{t}_{1 / 2}$ $\mathrm{t}=2 \times 3.20$ $\mathrm{t}=6.4 \mathrm{~h} \approx 6.38 \mathrm{~h}$
Manipal UGET-2010
NUCLEAR PHYSICS
147649
The number of active nuclei in two radioactive substances are in the ratio of $2: 3$ initially. If their half life's are one hour and two hours respectively, then the ratio of active nuclei after 6 hours is in the ratio of
1 $1: 1$
2 $1: 12$
3 $4: 3$
4 $12: 1$
Explanation:
B For sample first number of half lives, $\mathrm{n}_{1}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{1}=6$ For sample second number of half lives, $\mathrm{n}_{2}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{2}=3$ Ratio of initial amount $\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{1}}{\left(\mathrm{~N}_{\mathrm{o}}\right)_{2}}=\frac{2}{3}$ As we know, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ So, the ratio of active nuclei after 6 hours will be- $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3} \times \frac{(1 / 2)^{6}}{(1 / 2)^{3}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2 \times(2)^{3}}{3 \times(2)^{6}}=\frac{1}{12}$ So, $\quad \mathrm{N}_{1}: \mathrm{N}_{2}=1: 12$
147644
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. Calculate the minimum time after which it would be possible to work safely with this source.
147646
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
1 $\frac{1}{\mathrm{e}^{2}}$
2 $\frac{1}{\mathrm{e}}$
3 $\mathrm{e}$
4 $\mathrm{e}^{2}$
Explanation:
C Let, the initial number of atom $=\mathrm{N}_{0}$ The number of atom at an instant of time $=\mathrm{N}$ So, mean life $(\tau)=\frac{1}{\lambda}$ Given, Number of atom at an instant time $(\mathrm{t})=$ mean life $(\tau)$ According to radioactive law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \quad \ldots . . \text { (ii) } \quad(\therefore \mathrm{t}=1 / \lambda)$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\left(\lambda \times \frac{1}{\lambda}\right)}$ From equation (i) and equation (ii) $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$
Manipal UGET-2009
NUCLEAR PHYSICS
147647
The half-life of $\mathrm{At}^{215}$ is $100 \mu$ s. If a sample contains $215 \mathrm{mg}$ of $\mathrm{At}^{215}$, the activity of the sample initially is
1 $10^{2} \mathrm{~Bq}$
2 $3 \times 10^{10} \mathrm{~Bq}$
3 $4.17 \times 10^{24} \mathrm{~Bq}$
4 $1.16 \times 10^{5} \mathrm{~Bq}$
Explanation:
C Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=100 \mu \mathrm{s}=100 \times 10^{-6} \mathrm{~s}=10^{-4} \mathrm{~s}$ $\lambda=\frac{0.6931}{t_{1 / 2}}$ $\lambda=\frac{0.6931}{10^{-4}}=0.6931 \times 10^{4} \mathrm{~s}^{-1}$ Number of atoms in $215 \mathrm{mg}$ of sample $(\mathrm{N})-$ $=\frac{6.023 \times 10^{23} \times 215 \times 10^{-3}}{215}$ $\mathrm{~N}=6.023 \times 10^{20}$ Activity, $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=0.6931 \times 10^{4} \times 6.023 \times 10^{20}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=4.17 \times 10^{24} \mathrm{~Bq}$
Manipal UGET-2010
NUCLEAR PHYSICS
147648
Half-life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used ?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 4.18 day
4 1.2 day
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=3.20 \mathrm{~h}$ Percentage of substance is used $=75 \%$ So, Percentage of substance left $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=1-\frac{75}{100}$ $=25 \%$ $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=2$ $\mathrm{t}=2 \times \mathrm{t}_{1 / 2}$ $\mathrm{t}=2 \times 3.20$ $\mathrm{t}=6.4 \mathrm{~h} \approx 6.38 \mathrm{~h}$
Manipal UGET-2010
NUCLEAR PHYSICS
147649
The number of active nuclei in two radioactive substances are in the ratio of $2: 3$ initially. If their half life's are one hour and two hours respectively, then the ratio of active nuclei after 6 hours is in the ratio of
1 $1: 1$
2 $1: 12$
3 $4: 3$
4 $12: 1$
Explanation:
B For sample first number of half lives, $\mathrm{n}_{1}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{1}=6$ For sample second number of half lives, $\mathrm{n}_{2}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{2}=3$ Ratio of initial amount $\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{1}}{\left(\mathrm{~N}_{\mathrm{o}}\right)_{2}}=\frac{2}{3}$ As we know, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ So, the ratio of active nuclei after 6 hours will be- $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3} \times \frac{(1 / 2)^{6}}{(1 / 2)^{3}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2 \times(2)^{3}}{3 \times(2)^{6}}=\frac{1}{12}$ So, $\quad \mathrm{N}_{1}: \mathrm{N}_{2}=1: 12$
147644
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. Calculate the minimum time after which it would be possible to work safely with this source.
147646
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
1 $\frac{1}{\mathrm{e}^{2}}$
2 $\frac{1}{\mathrm{e}}$
3 $\mathrm{e}$
4 $\mathrm{e}^{2}$
Explanation:
C Let, the initial number of atom $=\mathrm{N}_{0}$ The number of atom at an instant of time $=\mathrm{N}$ So, mean life $(\tau)=\frac{1}{\lambda}$ Given, Number of atom at an instant time $(\mathrm{t})=$ mean life $(\tau)$ According to radioactive law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \quad \ldots . . \text { (ii) } \quad(\therefore \mathrm{t}=1 / \lambda)$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\left(\lambda \times \frac{1}{\lambda}\right)}$ From equation (i) and equation (ii) $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$
Manipal UGET-2009
NUCLEAR PHYSICS
147647
The half-life of $\mathrm{At}^{215}$ is $100 \mu$ s. If a sample contains $215 \mathrm{mg}$ of $\mathrm{At}^{215}$, the activity of the sample initially is
1 $10^{2} \mathrm{~Bq}$
2 $3 \times 10^{10} \mathrm{~Bq}$
3 $4.17 \times 10^{24} \mathrm{~Bq}$
4 $1.16 \times 10^{5} \mathrm{~Bq}$
Explanation:
C Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=100 \mu \mathrm{s}=100 \times 10^{-6} \mathrm{~s}=10^{-4} \mathrm{~s}$ $\lambda=\frac{0.6931}{t_{1 / 2}}$ $\lambda=\frac{0.6931}{10^{-4}}=0.6931 \times 10^{4} \mathrm{~s}^{-1}$ Number of atoms in $215 \mathrm{mg}$ of sample $(\mathrm{N})-$ $=\frac{6.023 \times 10^{23} \times 215 \times 10^{-3}}{215}$ $\mathrm{~N}=6.023 \times 10^{20}$ Activity, $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=0.6931 \times 10^{4} \times 6.023 \times 10^{20}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=4.17 \times 10^{24} \mathrm{~Bq}$
Manipal UGET-2010
NUCLEAR PHYSICS
147648
Half-life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used ?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 4.18 day
4 1.2 day
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=3.20 \mathrm{~h}$ Percentage of substance is used $=75 \%$ So, Percentage of substance left $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=1-\frac{75}{100}$ $=25 \%$ $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=2$ $\mathrm{t}=2 \times \mathrm{t}_{1 / 2}$ $\mathrm{t}=2 \times 3.20$ $\mathrm{t}=6.4 \mathrm{~h} \approx 6.38 \mathrm{~h}$
Manipal UGET-2010
NUCLEAR PHYSICS
147649
The number of active nuclei in two radioactive substances are in the ratio of $2: 3$ initially. If their half life's are one hour and two hours respectively, then the ratio of active nuclei after 6 hours is in the ratio of
1 $1: 1$
2 $1: 12$
3 $4: 3$
4 $12: 1$
Explanation:
B For sample first number of half lives, $\mathrm{n}_{1}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{1}=6$ For sample second number of half lives, $\mathrm{n}_{2}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{2}=3$ Ratio of initial amount $\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{1}}{\left(\mathrm{~N}_{\mathrm{o}}\right)_{2}}=\frac{2}{3}$ As we know, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ So, the ratio of active nuclei after 6 hours will be- $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3} \times \frac{(1 / 2)^{6}}{(1 / 2)^{3}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2 \times(2)^{3}}{3 \times(2)^{6}}=\frac{1}{12}$ So, $\quad \mathrm{N}_{1}: \mathrm{N}_{2}=1: 12$
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NUCLEAR PHYSICS
147644
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. Calculate the minimum time after which it would be possible to work safely with this source.
147646
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
1 $\frac{1}{\mathrm{e}^{2}}$
2 $\frac{1}{\mathrm{e}}$
3 $\mathrm{e}$
4 $\mathrm{e}^{2}$
Explanation:
C Let, the initial number of atom $=\mathrm{N}_{0}$ The number of atom at an instant of time $=\mathrm{N}$ So, mean life $(\tau)=\frac{1}{\lambda}$ Given, Number of atom at an instant time $(\mathrm{t})=$ mean life $(\tau)$ According to radioactive law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \quad \ldots . . \text { (ii) } \quad(\therefore \mathrm{t}=1 / \lambda)$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\left(\lambda \times \frac{1}{\lambda}\right)}$ From equation (i) and equation (ii) $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$
Manipal UGET-2009
NUCLEAR PHYSICS
147647
The half-life of $\mathrm{At}^{215}$ is $100 \mu$ s. If a sample contains $215 \mathrm{mg}$ of $\mathrm{At}^{215}$, the activity of the sample initially is
1 $10^{2} \mathrm{~Bq}$
2 $3 \times 10^{10} \mathrm{~Bq}$
3 $4.17 \times 10^{24} \mathrm{~Bq}$
4 $1.16 \times 10^{5} \mathrm{~Bq}$
Explanation:
C Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=100 \mu \mathrm{s}=100 \times 10^{-6} \mathrm{~s}=10^{-4} \mathrm{~s}$ $\lambda=\frac{0.6931}{t_{1 / 2}}$ $\lambda=\frac{0.6931}{10^{-4}}=0.6931 \times 10^{4} \mathrm{~s}^{-1}$ Number of atoms in $215 \mathrm{mg}$ of sample $(\mathrm{N})-$ $=\frac{6.023 \times 10^{23} \times 215 \times 10^{-3}}{215}$ $\mathrm{~N}=6.023 \times 10^{20}$ Activity, $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=0.6931 \times 10^{4} \times 6.023 \times 10^{20}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=4.17 \times 10^{24} \mathrm{~Bq}$
Manipal UGET-2010
NUCLEAR PHYSICS
147648
Half-life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used ?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 4.18 day
4 1.2 day
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=3.20 \mathrm{~h}$ Percentage of substance is used $=75 \%$ So, Percentage of substance left $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=1-\frac{75}{100}$ $=25 \%$ $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=2$ $\mathrm{t}=2 \times \mathrm{t}_{1 / 2}$ $\mathrm{t}=2 \times 3.20$ $\mathrm{t}=6.4 \mathrm{~h} \approx 6.38 \mathrm{~h}$
Manipal UGET-2010
NUCLEAR PHYSICS
147649
The number of active nuclei in two radioactive substances are in the ratio of $2: 3$ initially. If their half life's are one hour and two hours respectively, then the ratio of active nuclei after 6 hours is in the ratio of
1 $1: 1$
2 $1: 12$
3 $4: 3$
4 $12: 1$
Explanation:
B For sample first number of half lives, $\mathrm{n}_{1}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{1}=6$ For sample second number of half lives, $\mathrm{n}_{2}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{2}=3$ Ratio of initial amount $\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{1}}{\left(\mathrm{~N}_{\mathrm{o}}\right)_{2}}=\frac{2}{3}$ As we know, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ So, the ratio of active nuclei after 6 hours will be- $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3} \times \frac{(1 / 2)^{6}}{(1 / 2)^{3}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2 \times(2)^{3}}{3 \times(2)^{6}}=\frac{1}{12}$ So, $\quad \mathrm{N}_{1}: \mathrm{N}_{2}=1: 12$
147644
A freshly prepared radioactive source of halflife 2 h emits radiation of intensity which is 64 times the permissible safe level. Calculate the minimum time after which it would be possible to work safely with this source.
147646
In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at an instant of time equal to its mean life is
1 $\frac{1}{\mathrm{e}^{2}}$
2 $\frac{1}{\mathrm{e}}$
3 $\mathrm{e}$
4 $\mathrm{e}^{2}$
Explanation:
C Let, the initial number of atom $=\mathrm{N}_{0}$ The number of atom at an instant of time $=\mathrm{N}$ So, mean life $(\tau)=\frac{1}{\lambda}$ Given, Number of atom at an instant time $(\mathrm{t})=$ mean life $(\tau)$ According to radioactive law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}} \quad \ldots . . \text { (ii) } \quad(\therefore \mathrm{t}=1 / \lambda)$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\left(\lambda \times \frac{1}{\lambda}\right)}$ From equation (i) and equation (ii) $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}_{0}}{\mathrm{~N}}=\mathrm{e}$
Manipal UGET-2009
NUCLEAR PHYSICS
147647
The half-life of $\mathrm{At}^{215}$ is $100 \mu$ s. If a sample contains $215 \mathrm{mg}$ of $\mathrm{At}^{215}$, the activity of the sample initially is
1 $10^{2} \mathrm{~Bq}$
2 $3 \times 10^{10} \mathrm{~Bq}$
3 $4.17 \times 10^{24} \mathrm{~Bq}$
4 $1.16 \times 10^{5} \mathrm{~Bq}$
Explanation:
C Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=100 \mu \mathrm{s}=100 \times 10^{-6} \mathrm{~s}=10^{-4} \mathrm{~s}$ $\lambda=\frac{0.6931}{t_{1 / 2}}$ $\lambda=\frac{0.6931}{10^{-4}}=0.6931 \times 10^{4} \mathrm{~s}^{-1}$ Number of atoms in $215 \mathrm{mg}$ of sample $(\mathrm{N})-$ $=\frac{6.023 \times 10^{23} \times 215 \times 10^{-3}}{215}$ $\mathrm{~N}=6.023 \times 10^{20}$ Activity, $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=0.6931 \times 10^{4} \times 6.023 \times 10^{20}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=4.17 \times 10^{24} \mathrm{~Bq}$
Manipal UGET-2010
NUCLEAR PHYSICS
147648
Half-life of radioactive substance is $3.20 \mathrm{~h}$. What is the time taken for a $75 \%$ of substance to be used ?
1 $6.38 \mathrm{~h}$
2 $12 \mathrm{~h}$
3 4.18 day
4 1.2 day
Explanation:
A Given, Half life $\left(\mathrm{t}_{1 / 2}\right)=3.20 \mathrm{~h}$ Percentage of substance is used $=75 \%$ So, Percentage of substance left $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=1-\frac{75}{100}$ $=25 \%$ $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{25}{100}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=2$ $\mathrm{t}=2 \times \mathrm{t}_{1 / 2}$ $\mathrm{t}=2 \times 3.20$ $\mathrm{t}=6.4 \mathrm{~h} \approx 6.38 \mathrm{~h}$
Manipal UGET-2010
NUCLEAR PHYSICS
147649
The number of active nuclei in two radioactive substances are in the ratio of $2: 3$ initially. If their half life's are one hour and two hours respectively, then the ratio of active nuclei after 6 hours is in the ratio of
1 $1: 1$
2 $1: 12$
3 $4: 3$
4 $12: 1$
Explanation:
B For sample first number of half lives, $\mathrm{n}_{1}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{1}=6$ For sample second number of half lives, $\mathrm{n}_{2}=\frac{\mathrm{t}}{\mathrm{t}_{1 / 2}}=\frac{6}{2}=3$ Ratio of initial amount $\frac{\left(\mathrm{N}_{\mathrm{o}}\right)_{1}}{\left(\mathrm{~N}_{\mathrm{o}}\right)_{2}}=\frac{2}{3}$ As we know, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ So, the ratio of active nuclei after 6 hours will be- $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2}{3} \times \frac{(1 / 2)^{6}}{(1 / 2)^{3}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{2 \times(2)^{3}}{3 \times(2)^{6}}=\frac{1}{12}$ So, $\quad \mathrm{N}_{1}: \mathrm{N}_{2}=1: 12$