147866
The activity of a radioactive element decreased to one- third of the original activity $I_{0}$ in a period of nine years. After a further lapse of nine years its activity will be
1 $\mathrm{I}_{0}$
2 $\left(\frac{2}{3}\right) \mathrm{I}_{0}$
3 $\frac{\mathrm{I}_{0}}{9}$
4 $\frac{I_{0}}{6}$
Explanation:
C Given that, $\mathrm{t}=9$ years, $\mathrm{I}=\mathrm{I}_{\mathrm{o}} / 3$ If $I_{o}$ in initial activity $\mathrm{I}=\mathrm{I}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-9 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-9 \lambda}$ After 18 years, $I=I_{O} e^{-18 \lambda}$ $I=I_{0}\left(e^{-9 \lambda}\right)^{2}$ $I=I_{o} \times\left(\frac{1}{3}\right)^{2}$ $I=\frac{I_{0}}{9}$ Decreases to $\frac{1^{\text {th }}}{9}$ of original regards.
CG PET- 2010
NUCLEAR PHYSICS
147867
The half-life of ${ }^{215} \mathrm{At}$ is $100 \mu \mathrm{s}$. The time taken for the radioactivity of a sample of this nucleus to decay to $\frac{1}{16}$ th of its initial value is
1 $6.3 \mu \mathrm{s}$
2 $40 \mu \mathrm{s}$
3 $300 \mu \mathrm{s}$
4 $400 \mu \mathrm{s}$
Explanation:
D Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=100 \mu \mathrm{s}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ By using radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ On comparing both the side we get- $\frac{t}{100}=4$ $t=4 \times 100=400 \mu \mathrm{s}$
Manipal UGET-2016
NUCLEAR PHYSICS
147868
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then
A We know that, radioactive decay law- $-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\mathrm{R}=-\frac{\mathrm{dN}}{\mathrm{dt}}$ $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{R}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ At time $\mathrm{t}_{1}, \mathrm{R}_{1}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ Where, $\mathrm{R}_{0}=\lambda \mathrm{N}_{0}$ At time $t_{2}, R_{2}=R_{0} e^{-\lambda t_{2}}$ On dividing equation (ii) and (iii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{e}^{-\lambda \mathrm{t}_{1}}}{\mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
MHT-CET 2007
NUCLEAR PHYSICS
147869
If $8 \mathrm{~g}$ of a radioactive substance decays into 0.5 $\mathrm{g}$ in $\mathbf{1 ~} \mathrm{h}$, then the half-life of the substance is
1 $45 \mathrm{~min}$
2 $15 \mathrm{~min}$
3 $10 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
B Given that, Initial radioactive substance $\left(\mathrm{N}_{0}\right)=8 \mathrm{~g}$ Final radioactive substance $(\mathrm{N})=0.5 \mathrm{~g}$ Total time $(\mathrm{t})=1$ hour or $60 \mathrm{~min}$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $0.5=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{2}=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ On comparing both side, we get- $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{1 \times 60}{4}$ $\mathrm{~T}_{1 / 2}=15 \mathrm{~min}$
147866
The activity of a radioactive element decreased to one- third of the original activity $I_{0}$ in a period of nine years. After a further lapse of nine years its activity will be
1 $\mathrm{I}_{0}$
2 $\left(\frac{2}{3}\right) \mathrm{I}_{0}$
3 $\frac{\mathrm{I}_{0}}{9}$
4 $\frac{I_{0}}{6}$
Explanation:
C Given that, $\mathrm{t}=9$ years, $\mathrm{I}=\mathrm{I}_{\mathrm{o}} / 3$ If $I_{o}$ in initial activity $\mathrm{I}=\mathrm{I}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-9 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-9 \lambda}$ After 18 years, $I=I_{O} e^{-18 \lambda}$ $I=I_{0}\left(e^{-9 \lambda}\right)^{2}$ $I=I_{o} \times\left(\frac{1}{3}\right)^{2}$ $I=\frac{I_{0}}{9}$ Decreases to $\frac{1^{\text {th }}}{9}$ of original regards.
CG PET- 2010
NUCLEAR PHYSICS
147867
The half-life of ${ }^{215} \mathrm{At}$ is $100 \mu \mathrm{s}$. The time taken for the radioactivity of a sample of this nucleus to decay to $\frac{1}{16}$ th of its initial value is
1 $6.3 \mu \mathrm{s}$
2 $40 \mu \mathrm{s}$
3 $300 \mu \mathrm{s}$
4 $400 \mu \mathrm{s}$
Explanation:
D Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=100 \mu \mathrm{s}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ By using radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ On comparing both the side we get- $\frac{t}{100}=4$ $t=4 \times 100=400 \mu \mathrm{s}$
Manipal UGET-2016
NUCLEAR PHYSICS
147868
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then
A We know that, radioactive decay law- $-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\mathrm{R}=-\frac{\mathrm{dN}}{\mathrm{dt}}$ $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{R}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ At time $\mathrm{t}_{1}, \mathrm{R}_{1}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ Where, $\mathrm{R}_{0}=\lambda \mathrm{N}_{0}$ At time $t_{2}, R_{2}=R_{0} e^{-\lambda t_{2}}$ On dividing equation (ii) and (iii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{e}^{-\lambda \mathrm{t}_{1}}}{\mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
MHT-CET 2007
NUCLEAR PHYSICS
147869
If $8 \mathrm{~g}$ of a radioactive substance decays into 0.5 $\mathrm{g}$ in $\mathbf{1 ~} \mathrm{h}$, then the half-life of the substance is
1 $45 \mathrm{~min}$
2 $15 \mathrm{~min}$
3 $10 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
B Given that, Initial radioactive substance $\left(\mathrm{N}_{0}\right)=8 \mathrm{~g}$ Final radioactive substance $(\mathrm{N})=0.5 \mathrm{~g}$ Total time $(\mathrm{t})=1$ hour or $60 \mathrm{~min}$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $0.5=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{2}=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ On comparing both side, we get- $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{1 \times 60}{4}$ $\mathrm{~T}_{1 / 2}=15 \mathrm{~min}$
147866
The activity of a radioactive element decreased to one- third of the original activity $I_{0}$ in a period of nine years. After a further lapse of nine years its activity will be
1 $\mathrm{I}_{0}$
2 $\left(\frac{2}{3}\right) \mathrm{I}_{0}$
3 $\frac{\mathrm{I}_{0}}{9}$
4 $\frac{I_{0}}{6}$
Explanation:
C Given that, $\mathrm{t}=9$ years, $\mathrm{I}=\mathrm{I}_{\mathrm{o}} / 3$ If $I_{o}$ in initial activity $\mathrm{I}=\mathrm{I}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-9 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-9 \lambda}$ After 18 years, $I=I_{O} e^{-18 \lambda}$ $I=I_{0}\left(e^{-9 \lambda}\right)^{2}$ $I=I_{o} \times\left(\frac{1}{3}\right)^{2}$ $I=\frac{I_{0}}{9}$ Decreases to $\frac{1^{\text {th }}}{9}$ of original regards.
CG PET- 2010
NUCLEAR PHYSICS
147867
The half-life of ${ }^{215} \mathrm{At}$ is $100 \mu \mathrm{s}$. The time taken for the radioactivity of a sample of this nucleus to decay to $\frac{1}{16}$ th of its initial value is
1 $6.3 \mu \mathrm{s}$
2 $40 \mu \mathrm{s}$
3 $300 \mu \mathrm{s}$
4 $400 \mu \mathrm{s}$
Explanation:
D Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=100 \mu \mathrm{s}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ By using radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ On comparing both the side we get- $\frac{t}{100}=4$ $t=4 \times 100=400 \mu \mathrm{s}$
Manipal UGET-2016
NUCLEAR PHYSICS
147868
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then
A We know that, radioactive decay law- $-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\mathrm{R}=-\frac{\mathrm{dN}}{\mathrm{dt}}$ $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{R}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ At time $\mathrm{t}_{1}, \mathrm{R}_{1}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ Where, $\mathrm{R}_{0}=\lambda \mathrm{N}_{0}$ At time $t_{2}, R_{2}=R_{0} e^{-\lambda t_{2}}$ On dividing equation (ii) and (iii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{e}^{-\lambda \mathrm{t}_{1}}}{\mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
MHT-CET 2007
NUCLEAR PHYSICS
147869
If $8 \mathrm{~g}$ of a radioactive substance decays into 0.5 $\mathrm{g}$ in $\mathbf{1 ~} \mathrm{h}$, then the half-life of the substance is
1 $45 \mathrm{~min}$
2 $15 \mathrm{~min}$
3 $10 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
B Given that, Initial radioactive substance $\left(\mathrm{N}_{0}\right)=8 \mathrm{~g}$ Final radioactive substance $(\mathrm{N})=0.5 \mathrm{~g}$ Total time $(\mathrm{t})=1$ hour or $60 \mathrm{~min}$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $0.5=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{2}=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ On comparing both side, we get- $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{1 \times 60}{4}$ $\mathrm{~T}_{1 / 2}=15 \mathrm{~min}$
147866
The activity of a radioactive element decreased to one- third of the original activity $I_{0}$ in a period of nine years. After a further lapse of nine years its activity will be
1 $\mathrm{I}_{0}$
2 $\left(\frac{2}{3}\right) \mathrm{I}_{0}$
3 $\frac{\mathrm{I}_{0}}{9}$
4 $\frac{I_{0}}{6}$
Explanation:
C Given that, $\mathrm{t}=9$ years, $\mathrm{I}=\mathrm{I}_{\mathrm{o}} / 3$ If $I_{o}$ in initial activity $\mathrm{I}=\mathrm{I}_{0} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{I}_{\mathrm{o}}}{3}=\mathrm{I}_{\mathrm{o}} \mathrm{e}^{-9 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-9 \lambda}$ After 18 years, $I=I_{O} e^{-18 \lambda}$ $I=I_{0}\left(e^{-9 \lambda}\right)^{2}$ $I=I_{o} \times\left(\frac{1}{3}\right)^{2}$ $I=\frac{I_{0}}{9}$ Decreases to $\frac{1^{\text {th }}}{9}$ of original regards.
CG PET- 2010
NUCLEAR PHYSICS
147867
The half-life of ${ }^{215} \mathrm{At}$ is $100 \mu \mathrm{s}$. The time taken for the radioactivity of a sample of this nucleus to decay to $\frac{1}{16}$ th of its initial value is
1 $6.3 \mu \mathrm{s}$
2 $40 \mu \mathrm{s}$
3 $300 \mu \mathrm{s}$
4 $400 \mu \mathrm{s}$
Explanation:
D Given that, Half-life $\left(\mathrm{T}_{1 / 2}\right)=100 \mu \mathrm{s}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ By using radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{100}}$ $\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ On comparing both the side we get- $\frac{t}{100}=4$ $t=4 \times 100=400 \mu \mathrm{s}$
Manipal UGET-2016
NUCLEAR PHYSICS
147868
In a radioactive material the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$, it is $R_{2}$. If the decay constant of the material is $\lambda$, then
A We know that, radioactive decay law- $-\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $\mathrm{R}=-\frac{\mathrm{dN}}{\mathrm{dt}}$ $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{R}=\lambda \mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ At time $\mathrm{t}_{1}, \mathrm{R}_{1}=\mathrm{R}_{0} \mathrm{e}^{-\lambda \mathrm{t}_{1}}$ Where, $\mathrm{R}_{0}=\lambda \mathrm{N}_{0}$ At time $t_{2}, R_{2}=R_{0} e^{-\lambda t_{2}}$ On dividing equation (ii) and (iii), we get- $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{e}^{-\lambda \mathrm{t}_{1}}}{\mathrm{e}^{-\lambda \mathrm{t}_{2}}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$ $\mathrm{R}_{1}=\mathrm{R}_{2} \mathrm{e}^{-\lambda\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)}$
MHT-CET 2007
NUCLEAR PHYSICS
147869
If $8 \mathrm{~g}$ of a radioactive substance decays into 0.5 $\mathrm{g}$ in $\mathbf{1 ~} \mathrm{h}$, then the half-life of the substance is
1 $45 \mathrm{~min}$
2 $15 \mathrm{~min}$
3 $10 \mathrm{~min}$
4 $30 \mathrm{~min}$
Explanation:
B Given that, Initial radioactive substance $\left(\mathrm{N}_{0}\right)=8 \mathrm{~g}$ Final radioactive substance $(\mathrm{N})=0.5 \mathrm{~g}$ Total time $(\mathrm{t})=1$ hour or $60 \mathrm{~min}$ From the radioactive equation, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $0.5=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{2}=8 \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{t}{T_{1 / 2}}}$ On comparing both side, we get- $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{1 \times 60}{4}$ $\mathrm{~T}_{1 / 2}=15 \mathrm{~min}$
