147698
Activity of a radioactive sample decrease in $(1 / 3)^{\text {rd }}$ of its original value in 3 days. then, in 9 days its activity will becomes
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days - $\frac{1}{3}=\mathrm{e}^{-\lambda \times 3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$. Then, $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 3 \times 3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substitute the values of equation (i) in equation (ii), we get - $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\frac{1}{27}$ $\mathrm{~N}^{\prime}=\frac{\mathrm{N}_{0}}{27}$ Then, in 9 days its activity will become $\frac{1}{27}$ of the original value.
JCECE-2009
NUCLEAR PHYSICS
147700
Half-lives of two radioactive substances $A$ and $B$ are respectively $20 \mathrm{~min}$ and $\mathbf{4 0} \mathrm{min}$. Initially the samples of $A$ and $B$ have equal number of nuclei. After $80 \mathrm{~min}$ the ratio of remaining number of $A$ and $B$ nuclei is
1 $1: 16$
2 $4: 1$
3 $1: 4$
4 $1: 1$
Explanation:
C Given, Half life of substance $A\left(T_{1 / 2}\right)_{A}=20 \mathrm{~min}$ Half life of substance $B\left(T_{1 / 2}\right)_{B}=40 \mathrm{~min}$ Decay time $(\mathrm{t})=80 \mathrm{~min}$ Decay constant of substance A, $\lambda_{\mathrm{A}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{A}}}=\frac{\ln 2}{20}$ Decay constant of substance B, $\lambda_{\mathrm{B}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}}=\frac{\ln 2}{40}$ According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ For substance $\mathrm{A}$, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{A}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{20} \times 80}$ For substance $\mathrm{B}$, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{B}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{40 \times 80}}$ On dividing equation (i) \& (ii), we get - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times\left(\frac{1}{20}-\frac{1}{40}\right) \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times \frac{20}{800} \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 4}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{\mathrm{e}^{\ln 4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ So, $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JCECE-2007
NUCLEAR PHYSICS
147702
A radioactive element has half-life period 1600 yr. After $6400 \mathrm{yr}$, what part of element will remain?
1 $\frac{1}{4}$
2 $\frac{1}{8}$
3 $\frac{1}{16}$
4 $\frac{1}{2}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1600 \mathrm{yr}$ Decay time $(\mathrm{t})=6400 \mathrm{yr}$ The number of half lives $(n)=\frac{\text { time of decay }(t)}{\text { half } \operatorname{life}\left(T_{1 / 2}\right)}$ As radioactive decay is first-order kinetic half-life given is 1600 years in 6400 number of half life is - $\mathrm{n}=\frac{6400}{1600}=4$ Fraction of element remains after $\mathrm{n}$ half-life is given by $=\frac{1}{2^{n}}$ Hence, fraction of elements remains after 4 half-life, $\frac{1}{2^{4}}=\frac{1}{16}$
JCECE-2004
NUCLEAR PHYSICS
147703
The half-life of radium is $1600 \mathrm{yr}$. What is the mean life and disintegration constant of radium?
A The sum of half-lives of all atoms divided by the number of all atoms is called the mean-life $(\tau)$ of radioactive substance $\tau=\mathrm{T}_{1 / 2} / 0.693$ Where, $T_{1 / 2}$ is the half-life of radium $\tau=\frac{1600}{0.693} \mathrm{yr} \approx 2309 \mathrm{yr}$ Also mean-life of a radioactive substance is given by - $\tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{2309} / \mathrm{yr}$ So, the mean-life and disintegration constant are $\frac{1}{2309} / \mathrm{yr}$ and $2309 \mathrm{yr}$.
147698
Activity of a radioactive sample decrease in $(1 / 3)^{\text {rd }}$ of its original value in 3 days. then, in 9 days its activity will becomes
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days - $\frac{1}{3}=\mathrm{e}^{-\lambda \times 3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$. Then, $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 3 \times 3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substitute the values of equation (i) in equation (ii), we get - $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\frac{1}{27}$ $\mathrm{~N}^{\prime}=\frac{\mathrm{N}_{0}}{27}$ Then, in 9 days its activity will become $\frac{1}{27}$ of the original value.
JCECE-2009
NUCLEAR PHYSICS
147700
Half-lives of two radioactive substances $A$ and $B$ are respectively $20 \mathrm{~min}$ and $\mathbf{4 0} \mathrm{min}$. Initially the samples of $A$ and $B$ have equal number of nuclei. After $80 \mathrm{~min}$ the ratio of remaining number of $A$ and $B$ nuclei is
1 $1: 16$
2 $4: 1$
3 $1: 4$
4 $1: 1$
Explanation:
C Given, Half life of substance $A\left(T_{1 / 2}\right)_{A}=20 \mathrm{~min}$ Half life of substance $B\left(T_{1 / 2}\right)_{B}=40 \mathrm{~min}$ Decay time $(\mathrm{t})=80 \mathrm{~min}$ Decay constant of substance A, $\lambda_{\mathrm{A}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{A}}}=\frac{\ln 2}{20}$ Decay constant of substance B, $\lambda_{\mathrm{B}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}}=\frac{\ln 2}{40}$ According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ For substance $\mathrm{A}$, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{A}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{20} \times 80}$ For substance $\mathrm{B}$, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{B}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{40 \times 80}}$ On dividing equation (i) \& (ii), we get - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times\left(\frac{1}{20}-\frac{1}{40}\right) \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times \frac{20}{800} \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 4}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{\mathrm{e}^{\ln 4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ So, $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JCECE-2007
NUCLEAR PHYSICS
147702
A radioactive element has half-life period 1600 yr. After $6400 \mathrm{yr}$, what part of element will remain?
1 $\frac{1}{4}$
2 $\frac{1}{8}$
3 $\frac{1}{16}$
4 $\frac{1}{2}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1600 \mathrm{yr}$ Decay time $(\mathrm{t})=6400 \mathrm{yr}$ The number of half lives $(n)=\frac{\text { time of decay }(t)}{\text { half } \operatorname{life}\left(T_{1 / 2}\right)}$ As radioactive decay is first-order kinetic half-life given is 1600 years in 6400 number of half life is - $\mathrm{n}=\frac{6400}{1600}=4$ Fraction of element remains after $\mathrm{n}$ half-life is given by $=\frac{1}{2^{n}}$ Hence, fraction of elements remains after 4 half-life, $\frac{1}{2^{4}}=\frac{1}{16}$
JCECE-2004
NUCLEAR PHYSICS
147703
The half-life of radium is $1600 \mathrm{yr}$. What is the mean life and disintegration constant of radium?
A The sum of half-lives of all atoms divided by the number of all atoms is called the mean-life $(\tau)$ of radioactive substance $\tau=\mathrm{T}_{1 / 2} / 0.693$ Where, $T_{1 / 2}$ is the half-life of radium $\tau=\frac{1600}{0.693} \mathrm{yr} \approx 2309 \mathrm{yr}$ Also mean-life of a radioactive substance is given by - $\tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{2309} / \mathrm{yr}$ So, the mean-life and disintegration constant are $\frac{1}{2309} / \mathrm{yr}$ and $2309 \mathrm{yr}$.
147698
Activity of a radioactive sample decrease in $(1 / 3)^{\text {rd }}$ of its original value in 3 days. then, in 9 days its activity will becomes
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days - $\frac{1}{3}=\mathrm{e}^{-\lambda \times 3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$. Then, $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 3 \times 3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substitute the values of equation (i) in equation (ii), we get - $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\frac{1}{27}$ $\mathrm{~N}^{\prime}=\frac{\mathrm{N}_{0}}{27}$ Then, in 9 days its activity will become $\frac{1}{27}$ of the original value.
JCECE-2009
NUCLEAR PHYSICS
147700
Half-lives of two radioactive substances $A$ and $B$ are respectively $20 \mathrm{~min}$ and $\mathbf{4 0} \mathrm{min}$. Initially the samples of $A$ and $B$ have equal number of nuclei. After $80 \mathrm{~min}$ the ratio of remaining number of $A$ and $B$ nuclei is
1 $1: 16$
2 $4: 1$
3 $1: 4$
4 $1: 1$
Explanation:
C Given, Half life of substance $A\left(T_{1 / 2}\right)_{A}=20 \mathrm{~min}$ Half life of substance $B\left(T_{1 / 2}\right)_{B}=40 \mathrm{~min}$ Decay time $(\mathrm{t})=80 \mathrm{~min}$ Decay constant of substance A, $\lambda_{\mathrm{A}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{A}}}=\frac{\ln 2}{20}$ Decay constant of substance B, $\lambda_{\mathrm{B}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}}=\frac{\ln 2}{40}$ According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ For substance $\mathrm{A}$, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{A}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{20} \times 80}$ For substance $\mathrm{B}$, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{B}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{40 \times 80}}$ On dividing equation (i) \& (ii), we get - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times\left(\frac{1}{20}-\frac{1}{40}\right) \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times \frac{20}{800} \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 4}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{\mathrm{e}^{\ln 4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ So, $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JCECE-2007
NUCLEAR PHYSICS
147702
A radioactive element has half-life period 1600 yr. After $6400 \mathrm{yr}$, what part of element will remain?
1 $\frac{1}{4}$
2 $\frac{1}{8}$
3 $\frac{1}{16}$
4 $\frac{1}{2}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1600 \mathrm{yr}$ Decay time $(\mathrm{t})=6400 \mathrm{yr}$ The number of half lives $(n)=\frac{\text { time of decay }(t)}{\text { half } \operatorname{life}\left(T_{1 / 2}\right)}$ As radioactive decay is first-order kinetic half-life given is 1600 years in 6400 number of half life is - $\mathrm{n}=\frac{6400}{1600}=4$ Fraction of element remains after $\mathrm{n}$ half-life is given by $=\frac{1}{2^{n}}$ Hence, fraction of elements remains after 4 half-life, $\frac{1}{2^{4}}=\frac{1}{16}$
JCECE-2004
NUCLEAR PHYSICS
147703
The half-life of radium is $1600 \mathrm{yr}$. What is the mean life and disintegration constant of radium?
A The sum of half-lives of all atoms divided by the number of all atoms is called the mean-life $(\tau)$ of radioactive substance $\tau=\mathrm{T}_{1 / 2} / 0.693$ Where, $T_{1 / 2}$ is the half-life of radium $\tau=\frac{1600}{0.693} \mathrm{yr} \approx 2309 \mathrm{yr}$ Also mean-life of a radioactive substance is given by - $\tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{2309} / \mathrm{yr}$ So, the mean-life and disintegration constant are $\frac{1}{2309} / \mathrm{yr}$ and $2309 \mathrm{yr}$.
147698
Activity of a radioactive sample decrease in $(1 / 3)^{\text {rd }}$ of its original value in 3 days. then, in 9 days its activity will becomes
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days - $\frac{1}{3}=\mathrm{e}^{-\lambda \times 3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$. Then, $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 3 \times 3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substitute the values of equation (i) in equation (ii), we get - $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\frac{1}{27}$ $\mathrm{~N}^{\prime}=\frac{\mathrm{N}_{0}}{27}$ Then, in 9 days its activity will become $\frac{1}{27}$ of the original value.
JCECE-2009
NUCLEAR PHYSICS
147700
Half-lives of two radioactive substances $A$ and $B$ are respectively $20 \mathrm{~min}$ and $\mathbf{4 0} \mathrm{min}$. Initially the samples of $A$ and $B$ have equal number of nuclei. After $80 \mathrm{~min}$ the ratio of remaining number of $A$ and $B$ nuclei is
1 $1: 16$
2 $4: 1$
3 $1: 4$
4 $1: 1$
Explanation:
C Given, Half life of substance $A\left(T_{1 / 2}\right)_{A}=20 \mathrm{~min}$ Half life of substance $B\left(T_{1 / 2}\right)_{B}=40 \mathrm{~min}$ Decay time $(\mathrm{t})=80 \mathrm{~min}$ Decay constant of substance A, $\lambda_{\mathrm{A}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{A}}}=\frac{\ln 2}{20}$ Decay constant of substance B, $\lambda_{\mathrm{B}}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{\mathrm{B}}}=\frac{\ln 2}{40}$ According to radioactive decay law - $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ For substance $\mathrm{A}$, $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{A}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{20} \times 80}$ For substance $\mathrm{B}$, $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{-\lambda_{\mathrm{B}} \mathrm{t}}$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{0} \mathrm{e}^{\frac{\ln 2}{40 \times 80}}$ On dividing equation (i) \& (ii), we get - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times\left(\frac{1}{20}-\frac{1}{40}\right) \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 2 \times \frac{20}{800} \times 80}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\mathrm{e}^{-\ln 4}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{\mathrm{e}^{\ln 4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ So, $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JCECE-2007
NUCLEAR PHYSICS
147702
A radioactive element has half-life period 1600 yr. After $6400 \mathrm{yr}$, what part of element will remain?
1 $\frac{1}{4}$
2 $\frac{1}{8}$
3 $\frac{1}{16}$
4 $\frac{1}{2}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1600 \mathrm{yr}$ Decay time $(\mathrm{t})=6400 \mathrm{yr}$ The number of half lives $(n)=\frac{\text { time of decay }(t)}{\text { half } \operatorname{life}\left(T_{1 / 2}\right)}$ As radioactive decay is first-order kinetic half-life given is 1600 years in 6400 number of half life is - $\mathrm{n}=\frac{6400}{1600}=4$ Fraction of element remains after $\mathrm{n}$ half-life is given by $=\frac{1}{2^{n}}$ Hence, fraction of elements remains after 4 half-life, $\frac{1}{2^{4}}=\frac{1}{16}$
JCECE-2004
NUCLEAR PHYSICS
147703
The half-life of radium is $1600 \mathrm{yr}$. What is the mean life and disintegration constant of radium?
A The sum of half-lives of all atoms divided by the number of all atoms is called the mean-life $(\tau)$ of radioactive substance $\tau=\mathrm{T}_{1 / 2} / 0.693$ Where, $T_{1 / 2}$ is the half-life of radium $\tau=\frac{1600}{0.693} \mathrm{yr} \approx 2309 \mathrm{yr}$ Also mean-life of a radioactive substance is given by - $\tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{2309} / \mathrm{yr}$ So, the mean-life and disintegration constant are $\frac{1}{2309} / \mathrm{yr}$ and $2309 \mathrm{yr}$.
