147802
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\lambda$
2 $\frac{1}{2} \lambda$
3 $\frac{1}{4 \lambda}$
4 $\frac{\mathrm{e}}{\lambda}$
Explanation:
C Considering first order decay process and same number of initial number of atom. $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\text { And }\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}_{0} \mathrm{e}^{-\lambda \lambda t}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-5 \lambda \mathrm{t}} \times \mathrm{e}^{+\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-4 \lambda t}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-4 \lambda \mathrm{t}}$ $\mathrm{e}^{1}=\mathrm{e}^{4 \lambda \mathrm{t}}$ On comparing both side we get - $4 \lambda t=1$ $\mathrm{t}=\frac{1}{4 \lambda}$
COMEDK- 2020
NUCLEAR PHYSICS
147803
The half life of a radioactive substance is 7.5 seconds. The fraction of substance left after one minute is
1 $1 / 16$
2 $1 / 64$
3 $1 / 128$
4 $1 / 256$
Explanation:
D Given, Half life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=7.5$ second Total time of decay $(\mathrm{t})=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{60 / 7.5}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{256}$
UPSEE 2020
NUCLEAR PHYSICS
147804
At given instant of time there are $25 \%$ undecayed unclei in a sample. After $10 \mathrm{~s}$ number of undecayed nuclei reduce to $12.5 \%$. What is mean life of the Nuclei? (nearly)
1 $15 \mathrm{~s}$
2 $22 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $20 \mathrm{~s}$
Explanation:
A Given, half-life $=10 \mathrm{sec}$ We know that, Decay constant $(\lambda)=\frac{0.693}{10}=0.0693$ So, Mean life $(\tau)=\frac{1}{\lambda}$ $\tau=\frac{1}{0.0693}$ $\tau=14.43$ $\tau \approx 15 \mathrm{sec}$
AP EAMCET (Medical)-07.10.2020
NUCLEAR PHYSICS
147805
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 that, Half-lives $\left(\mathrm{T}_{\mathrm{A}}\right)=20 \mathrm{~min}$ $\left(\mathrm{T}_{\mathrm{B}}\right)=40 \mathrm{~min}$ Total time taken $(\mathrm{t})=80 \mathrm{~min}$ We know that, $\mathrm{n}_{\mathrm{A}}=\frac{\mathrm{t}}{\mathrm{T}_{\mathrm{A}}}=\frac{80}{20}=4$ $\text { So, } \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{o}}}{16}$ Similarly, $\mathrm{n}_{\mathrm{B}}=\frac{80}{40}=2$ $\text { So, } \mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}_{\mathrm{B}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ The ratio of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ is - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{16}}{\frac{\mathrm{N}_{\mathrm{o}}}{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JIPMER-2008
NUCLEAR PHYSICS
147806
A radio-active elements has half-life of 15 years. What is the fraction that will decay in $\mathbf{3 0}$ years?
1 0.25
2 0.5
3 0.75
4 0.85
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=15$ years Total time $(\mathrm{t})=30$ year We know that, radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T} / 2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=0.25$ Fraction for undecay $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-0.25$ $=0.75$
147802
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\lambda$
2 $\frac{1}{2} \lambda$
3 $\frac{1}{4 \lambda}$
4 $\frac{\mathrm{e}}{\lambda}$
Explanation:
C Considering first order decay process and same number of initial number of atom. $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\text { And }\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}_{0} \mathrm{e}^{-\lambda \lambda t}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-5 \lambda \mathrm{t}} \times \mathrm{e}^{+\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-4 \lambda t}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-4 \lambda \mathrm{t}}$ $\mathrm{e}^{1}=\mathrm{e}^{4 \lambda \mathrm{t}}$ On comparing both side we get - $4 \lambda t=1$ $\mathrm{t}=\frac{1}{4 \lambda}$
COMEDK- 2020
NUCLEAR PHYSICS
147803
The half life of a radioactive substance is 7.5 seconds. The fraction of substance left after one minute is
1 $1 / 16$
2 $1 / 64$
3 $1 / 128$
4 $1 / 256$
Explanation:
D Given, Half life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=7.5$ second Total time of decay $(\mathrm{t})=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{60 / 7.5}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{256}$
UPSEE 2020
NUCLEAR PHYSICS
147804
At given instant of time there are $25 \%$ undecayed unclei in a sample. After $10 \mathrm{~s}$ number of undecayed nuclei reduce to $12.5 \%$. What is mean life of the Nuclei? (nearly)
1 $15 \mathrm{~s}$
2 $22 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $20 \mathrm{~s}$
Explanation:
A Given, half-life $=10 \mathrm{sec}$ We know that, Decay constant $(\lambda)=\frac{0.693}{10}=0.0693$ So, Mean life $(\tau)=\frac{1}{\lambda}$ $\tau=\frac{1}{0.0693}$ $\tau=14.43$ $\tau \approx 15 \mathrm{sec}$
AP EAMCET (Medical)-07.10.2020
NUCLEAR PHYSICS
147805
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 that, Half-lives $\left(\mathrm{T}_{\mathrm{A}}\right)=20 \mathrm{~min}$ $\left(\mathrm{T}_{\mathrm{B}}\right)=40 \mathrm{~min}$ Total time taken $(\mathrm{t})=80 \mathrm{~min}$ We know that, $\mathrm{n}_{\mathrm{A}}=\frac{\mathrm{t}}{\mathrm{T}_{\mathrm{A}}}=\frac{80}{20}=4$ $\text { So, } \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{o}}}{16}$ Similarly, $\mathrm{n}_{\mathrm{B}}=\frac{80}{40}=2$ $\text { So, } \mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}_{\mathrm{B}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ The ratio of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ is - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{16}}{\frac{\mathrm{N}_{\mathrm{o}}}{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JIPMER-2008
NUCLEAR PHYSICS
147806
A radio-active elements has half-life of 15 years. What is the fraction that will decay in $\mathbf{3 0}$ years?
1 0.25
2 0.5
3 0.75
4 0.85
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=15$ years Total time $(\mathrm{t})=30$ year We know that, radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T} / 2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=0.25$ Fraction for undecay $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-0.25$ $=0.75$
147802
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\lambda$
2 $\frac{1}{2} \lambda$
3 $\frac{1}{4 \lambda}$
4 $\frac{\mathrm{e}}{\lambda}$
Explanation:
C Considering first order decay process and same number of initial number of atom. $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\text { And }\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}_{0} \mathrm{e}^{-\lambda \lambda t}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-5 \lambda \mathrm{t}} \times \mathrm{e}^{+\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-4 \lambda t}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-4 \lambda \mathrm{t}}$ $\mathrm{e}^{1}=\mathrm{e}^{4 \lambda \mathrm{t}}$ On comparing both side we get - $4 \lambda t=1$ $\mathrm{t}=\frac{1}{4 \lambda}$
COMEDK- 2020
NUCLEAR PHYSICS
147803
The half life of a radioactive substance is 7.5 seconds. The fraction of substance left after one minute is
1 $1 / 16$
2 $1 / 64$
3 $1 / 128$
4 $1 / 256$
Explanation:
D Given, Half life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=7.5$ second Total time of decay $(\mathrm{t})=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{60 / 7.5}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{256}$
UPSEE 2020
NUCLEAR PHYSICS
147804
At given instant of time there are $25 \%$ undecayed unclei in a sample. After $10 \mathrm{~s}$ number of undecayed nuclei reduce to $12.5 \%$. What is mean life of the Nuclei? (nearly)
1 $15 \mathrm{~s}$
2 $22 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $20 \mathrm{~s}$
Explanation:
A Given, half-life $=10 \mathrm{sec}$ We know that, Decay constant $(\lambda)=\frac{0.693}{10}=0.0693$ So, Mean life $(\tau)=\frac{1}{\lambda}$ $\tau=\frac{1}{0.0693}$ $\tau=14.43$ $\tau \approx 15 \mathrm{sec}$
AP EAMCET (Medical)-07.10.2020
NUCLEAR PHYSICS
147805
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 that, Half-lives $\left(\mathrm{T}_{\mathrm{A}}\right)=20 \mathrm{~min}$ $\left(\mathrm{T}_{\mathrm{B}}\right)=40 \mathrm{~min}$ Total time taken $(\mathrm{t})=80 \mathrm{~min}$ We know that, $\mathrm{n}_{\mathrm{A}}=\frac{\mathrm{t}}{\mathrm{T}_{\mathrm{A}}}=\frac{80}{20}=4$ $\text { So, } \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{o}}}{16}$ Similarly, $\mathrm{n}_{\mathrm{B}}=\frac{80}{40}=2$ $\text { So, } \mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}_{\mathrm{B}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ The ratio of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ is - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{16}}{\frac{\mathrm{N}_{\mathrm{o}}}{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JIPMER-2008
NUCLEAR PHYSICS
147806
A radio-active elements has half-life of 15 years. What is the fraction that will decay in $\mathbf{3 0}$ years?
1 0.25
2 0.5
3 0.75
4 0.85
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=15$ years Total time $(\mathrm{t})=30$ year We know that, radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T} / 2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=0.25$ Fraction for undecay $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-0.25$ $=0.75$
147802
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\lambda$
2 $\frac{1}{2} \lambda$
3 $\frac{1}{4 \lambda}$
4 $\frac{\mathrm{e}}{\lambda}$
Explanation:
C Considering first order decay process and same number of initial number of atom. $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\text { And }\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}_{0} \mathrm{e}^{-\lambda \lambda t}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-5 \lambda \mathrm{t}} \times \mathrm{e}^{+\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-4 \lambda t}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-4 \lambda \mathrm{t}}$ $\mathrm{e}^{1}=\mathrm{e}^{4 \lambda \mathrm{t}}$ On comparing both side we get - $4 \lambda t=1$ $\mathrm{t}=\frac{1}{4 \lambda}$
COMEDK- 2020
NUCLEAR PHYSICS
147803
The half life of a radioactive substance is 7.5 seconds. The fraction of substance left after one minute is
1 $1 / 16$
2 $1 / 64$
3 $1 / 128$
4 $1 / 256$
Explanation:
D Given, Half life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=7.5$ second Total time of decay $(\mathrm{t})=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{60 / 7.5}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{256}$
UPSEE 2020
NUCLEAR PHYSICS
147804
At given instant of time there are $25 \%$ undecayed unclei in a sample. After $10 \mathrm{~s}$ number of undecayed nuclei reduce to $12.5 \%$. What is mean life of the Nuclei? (nearly)
1 $15 \mathrm{~s}$
2 $22 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $20 \mathrm{~s}$
Explanation:
A Given, half-life $=10 \mathrm{sec}$ We know that, Decay constant $(\lambda)=\frac{0.693}{10}=0.0693$ So, Mean life $(\tau)=\frac{1}{\lambda}$ $\tau=\frac{1}{0.0693}$ $\tau=14.43$ $\tau \approx 15 \mathrm{sec}$
AP EAMCET (Medical)-07.10.2020
NUCLEAR PHYSICS
147805
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 that, Half-lives $\left(\mathrm{T}_{\mathrm{A}}\right)=20 \mathrm{~min}$ $\left(\mathrm{T}_{\mathrm{B}}\right)=40 \mathrm{~min}$ Total time taken $(\mathrm{t})=80 \mathrm{~min}$ We know that, $\mathrm{n}_{\mathrm{A}}=\frac{\mathrm{t}}{\mathrm{T}_{\mathrm{A}}}=\frac{80}{20}=4$ $\text { So, } \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{o}}}{16}$ Similarly, $\mathrm{n}_{\mathrm{B}}=\frac{80}{40}=2$ $\text { So, } \mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}_{\mathrm{B}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ The ratio of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ is - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{16}}{\frac{\mathrm{N}_{\mathrm{o}}}{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JIPMER-2008
NUCLEAR PHYSICS
147806
A radio-active elements has half-life of 15 years. What is the fraction that will decay in $\mathbf{3 0}$ years?
1 0.25
2 0.5
3 0.75
4 0.85
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=15$ years Total time $(\mathrm{t})=30$ year We know that, radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T} / 2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=0.25$ Fraction for undecay $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-0.25$ $=0.75$
147802
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\lambda$
2 $\frac{1}{2} \lambda$
3 $\frac{1}{4 \lambda}$
4 $\frac{\mathrm{e}}{\lambda}$
Explanation:
C Considering first order decay process and same number of initial number of atom. $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ $\text { And }\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}_{0} \mathrm{e}^{-\lambda \lambda t}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-5 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-5 \lambda \mathrm{t}} \times \mathrm{e}^{+\lambda \mathrm{t}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-4 \lambda t}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-4 \lambda \mathrm{t}}$ $\mathrm{e}^{1}=\mathrm{e}^{4 \lambda \mathrm{t}}$ On comparing both side we get - $4 \lambda t=1$ $\mathrm{t}=\frac{1}{4 \lambda}$
COMEDK- 2020
NUCLEAR PHYSICS
147803
The half life of a radioactive substance is 7.5 seconds. The fraction of substance left after one minute is
1 $1 / 16$
2 $1 / 64$
3 $1 / 128$
4 $1 / 256$
Explanation:
D Given, Half life of radioactive substance $\left(\mathrm{T}_{1 / 2}\right)=7.5$ second Total time of decay $(\mathrm{t})=1 \mathrm{~min}=60 \mathrm{sec}$ We know that, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{60 / 7.5}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{256}$
UPSEE 2020
NUCLEAR PHYSICS
147804
At given instant of time there are $25 \%$ undecayed unclei in a sample. After $10 \mathrm{~s}$ number of undecayed nuclei reduce to $12.5 \%$. What is mean life of the Nuclei? (nearly)
1 $15 \mathrm{~s}$
2 $22 \mathrm{~s}$
3 $10 \mathrm{~s}$
4 $20 \mathrm{~s}$
Explanation:
A Given, half-life $=10 \mathrm{sec}$ We know that, Decay constant $(\lambda)=\frac{0.693}{10}=0.0693$ So, Mean life $(\tau)=\frac{1}{\lambda}$ $\tau=\frac{1}{0.0693}$ $\tau=14.43$ $\tau \approx 15 \mathrm{sec}$
AP EAMCET (Medical)-07.10.2020
NUCLEAR PHYSICS
147805
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 that, Half-lives $\left(\mathrm{T}_{\mathrm{A}}\right)=20 \mathrm{~min}$ $\left(\mathrm{T}_{\mathrm{B}}\right)=40 \mathrm{~min}$ Total time taken $(\mathrm{t})=80 \mathrm{~min}$ We know that, $\mathrm{n}_{\mathrm{A}}=\frac{\mathrm{t}}{\mathrm{T}_{\mathrm{A}}}=\frac{80}{20}=4$ $\text { So, } \mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{4}$ $\mathrm{~N}_{\mathrm{A}}=\frac{\mathrm{N}_{\mathrm{o}}}{16}$ Similarly, $\mathrm{n}_{\mathrm{B}}=\frac{80}{40}=2$ $\text { So, } \mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}_{\mathrm{B}}=\frac{\mathrm{N}_{\mathrm{o}}}{4}$ The ratio of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ is - $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{\frac{\mathrm{N}_{\mathrm{o}}}{16}}{\frac{\mathrm{N}_{\mathrm{o}}}{4}}$ $\frac{\mathrm{N}_{\mathrm{A}}}{\mathrm{N}_{\mathrm{B}}}=\frac{1}{4}$ $\mathrm{N}_{\mathrm{A}}: \mathrm{N}_{\mathrm{B}}=1: 4$
JIPMER-2008
NUCLEAR PHYSICS
147806
A radio-active elements has half-life of 15 years. What is the fraction that will decay in $\mathbf{3 0}$ years?
1 0.25
2 0.5
3 0.75
4 0.85
Explanation:
C Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=15$ years Total time $(\mathrm{t})=30$ year We know that, radioactive equation- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T} / 2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=0.25$ Fraction for undecay $=1-\frac{\mathrm{N}}{\mathrm{N}_{0}}$ $=1-0.25$ $=0.75$
