147826
If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $\mathbf{7 5 \%}$.
1 27.7 years
2 57.7 years
3 60 years
4 87 years
Explanation:
A Given that, Decay constant $(\lambda)=0.05 /$ year When $75 \%$ decay, $\mathrm{N}=\mathrm{N}_{0}-\frac{3 \mathrm{~N}_{0}}{4}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{4}$ We know that, $\ln \frac{\mathrm{N}}{\mathrm{N}_{0}}=-\lambda \mathrm{t}$ $\ln \left(\frac{1}{4}\right)=-\lambda \mathrm{t}$ $-1.386=-\lambda \mathrm{t}$ $\lambda=\frac{1.386}{\mathrm{t}}=\frac{1.386}{0.05}$ $\mathrm{t}=27.72 \text { years }$
AIIMS-27.05.2018(M)
NUCLEAR PHYSICS
147827
Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $\mathbf{5 0}$ days is equal to that of $1.2 \times 10^{8}$ number of undecayed nuclei of some other material with half-life of 60 days. Reason: Radioactivity is proportional to half -life.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
147828
Radioactive material $A$ has decay constant $8 \lambda$ and material $B$ has decay constant $\lambda$. Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{\mathrm{e}}$ ?
1 $\frac{1}{\lambda}$
2 $\frac{1}{7 \lambda}$
3 $\frac{1}{8 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
B Considering first order decay process and same number of initial number of atom. We know that, $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\text { and }\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}$ $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ on dividing eqaution (i) by (ii) we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-7 \lambda t}$ $\mathrm{e}^{-1}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ On comparing both side we get - $-7 \lambda \mathrm{t}=-1$ $\mathrm{t}=\frac{1}{7 \lambda}$
NEET- 2017
NUCLEAR PHYSICS
147829
The half-life of a radioactive isotope ' $X$ ' is 20 years. It decays to antoher stable element ' $Y$ '. The number of atoms in ' $X$ ' and ' $Y$ ' are found to be in the ratio. 1:7 in a rock. The age of the rock is
1 40 years
2 60 years
3 80 years
4 100 Years
Explanation:
B Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=20$ years According to question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}-\mathrm{N}}=\frac{1}{7}$ $7 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $8 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \Rightarrow \quad\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{20}}$ On comparing both side we get- $\frac{t}{20}=3$ $t=20 \times 3$ $t=60 \text { years }$
147826
If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $\mathbf{7 5 \%}$.
1 27.7 years
2 57.7 years
3 60 years
4 87 years
Explanation:
A Given that, Decay constant $(\lambda)=0.05 /$ year When $75 \%$ decay, $\mathrm{N}=\mathrm{N}_{0}-\frac{3 \mathrm{~N}_{0}}{4}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{4}$ We know that, $\ln \frac{\mathrm{N}}{\mathrm{N}_{0}}=-\lambda \mathrm{t}$ $\ln \left(\frac{1}{4}\right)=-\lambda \mathrm{t}$ $-1.386=-\lambda \mathrm{t}$ $\lambda=\frac{1.386}{\mathrm{t}}=\frac{1.386}{0.05}$ $\mathrm{t}=27.72 \text { years }$
AIIMS-27.05.2018(M)
NUCLEAR PHYSICS
147827
Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $\mathbf{5 0}$ days is equal to that of $1.2 \times 10^{8}$ number of undecayed nuclei of some other material with half-life of 60 days. Reason: Radioactivity is proportional to half -life.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
147828
Radioactive material $A$ has decay constant $8 \lambda$ and material $B$ has decay constant $\lambda$. Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{\mathrm{e}}$ ?
1 $\frac{1}{\lambda}$
2 $\frac{1}{7 \lambda}$
3 $\frac{1}{8 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
B Considering first order decay process and same number of initial number of atom. We know that, $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\text { and }\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}$ $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ on dividing eqaution (i) by (ii) we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-7 \lambda t}$ $\mathrm{e}^{-1}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ On comparing both side we get - $-7 \lambda \mathrm{t}=-1$ $\mathrm{t}=\frac{1}{7 \lambda}$
NEET- 2017
NUCLEAR PHYSICS
147829
The half-life of a radioactive isotope ' $X$ ' is 20 years. It decays to antoher stable element ' $Y$ '. The number of atoms in ' $X$ ' and ' $Y$ ' are found to be in the ratio. 1:7 in a rock. The age of the rock is
1 40 years
2 60 years
3 80 years
4 100 Years
Explanation:
B Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=20$ years According to question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}-\mathrm{N}}=\frac{1}{7}$ $7 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $8 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \Rightarrow \quad\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{20}}$ On comparing both side we get- $\frac{t}{20}=3$ $t=20 \times 3$ $t=60 \text { years }$
147826
If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $\mathbf{7 5 \%}$.
1 27.7 years
2 57.7 years
3 60 years
4 87 years
Explanation:
A Given that, Decay constant $(\lambda)=0.05 /$ year When $75 \%$ decay, $\mathrm{N}=\mathrm{N}_{0}-\frac{3 \mathrm{~N}_{0}}{4}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{4}$ We know that, $\ln \frac{\mathrm{N}}{\mathrm{N}_{0}}=-\lambda \mathrm{t}$ $\ln \left(\frac{1}{4}\right)=-\lambda \mathrm{t}$ $-1.386=-\lambda \mathrm{t}$ $\lambda=\frac{1.386}{\mathrm{t}}=\frac{1.386}{0.05}$ $\mathrm{t}=27.72 \text { years }$
AIIMS-27.05.2018(M)
NUCLEAR PHYSICS
147827
Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $\mathbf{5 0}$ days is equal to that of $1.2 \times 10^{8}$ number of undecayed nuclei of some other material with half-life of 60 days. Reason: Radioactivity is proportional to half -life.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
147828
Radioactive material $A$ has decay constant $8 \lambda$ and material $B$ has decay constant $\lambda$. Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{\mathrm{e}}$ ?
1 $\frac{1}{\lambda}$
2 $\frac{1}{7 \lambda}$
3 $\frac{1}{8 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
B Considering first order decay process and same number of initial number of atom. We know that, $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\text { and }\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}$ $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ on dividing eqaution (i) by (ii) we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-7 \lambda t}$ $\mathrm{e}^{-1}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ On comparing both side we get - $-7 \lambda \mathrm{t}=-1$ $\mathrm{t}=\frac{1}{7 \lambda}$
NEET- 2017
NUCLEAR PHYSICS
147829
The half-life of a radioactive isotope ' $X$ ' is 20 years. It decays to antoher stable element ' $Y$ '. The number of atoms in ' $X$ ' and ' $Y$ ' are found to be in the ratio. 1:7 in a rock. The age of the rock is
1 40 years
2 60 years
3 80 years
4 100 Years
Explanation:
B Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=20$ years According to question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}-\mathrm{N}}=\frac{1}{7}$ $7 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $8 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \Rightarrow \quad\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{20}}$ On comparing both side we get- $\frac{t}{20}=3$ $t=20 \times 3$ $t=60 \text { years }$
147826
If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $\mathbf{7 5 \%}$.
1 27.7 years
2 57.7 years
3 60 years
4 87 years
Explanation:
A Given that, Decay constant $(\lambda)=0.05 /$ year When $75 \%$ decay, $\mathrm{N}=\mathrm{N}_{0}-\frac{3 \mathrm{~N}_{0}}{4}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{4}$ We know that, $\ln \frac{\mathrm{N}}{\mathrm{N}_{0}}=-\lambda \mathrm{t}$ $\ln \left(\frac{1}{4}\right)=-\lambda \mathrm{t}$ $-1.386=-\lambda \mathrm{t}$ $\lambda=\frac{1.386}{\mathrm{t}}=\frac{1.386}{0.05}$ $\mathrm{t}=27.72 \text { years }$
AIIMS-27.05.2018(M)
NUCLEAR PHYSICS
147827
Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $\mathbf{5 0}$ days is equal to that of $1.2 \times 10^{8}$ number of undecayed nuclei of some other material with half-life of 60 days. Reason: Radioactivity is proportional to half -life.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
147828
Radioactive material $A$ has decay constant $8 \lambda$ and material $B$ has decay constant $\lambda$. Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{\mathrm{e}}$ ?
1 $\frac{1}{\lambda}$
2 $\frac{1}{7 \lambda}$
3 $\frac{1}{8 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
B Considering first order decay process and same number of initial number of atom. We know that, $\text { So, }\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\text { and }\mathrm{N}_{1}=\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}$ $\mathrm{N}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}$ on dividing eqaution (i) by (ii) we get - $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{\mathrm{N}_{0} \mathrm{e}^{-8 \lambda \mathrm{t}}}{\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}}$ $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ Given that, $\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}}=\frac{1}{\mathrm{e}}$ $\text { Hence, } \frac{1}{\mathrm{e}}=\mathrm{e}^{-7 \lambda t}$ $\mathrm{e}^{-1}=\mathrm{e}^{-7 \lambda \mathrm{t}}$ On comparing both side we get - $-7 \lambda \mathrm{t}=-1$ $\mathrm{t}=\frac{1}{7 \lambda}$
NEET- 2017
NUCLEAR PHYSICS
147829
The half-life of a radioactive isotope ' $X$ ' is 20 years. It decays to antoher stable element ' $Y$ '. The number of atoms in ' $X$ ' and ' $Y$ ' are found to be in the ratio. 1:7 in a rock. The age of the rock is
1 40 years
2 60 years
3 80 years
4 100 Years
Explanation:
B Given that, half-life $\left(\mathrm{T}_{1 / 2}\right)=20$ years According to question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}-\mathrm{N}}=\frac{1}{7}$ $7 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}-\mathrm{N}$ $8 \mathrm{~N}=\mathrm{N}_{\mathrm{o}}$ $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{1}{8}$ We know that, radioactive equation, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad\left(\because \mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}\right)$ $\frac{1}{8}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \Rightarrow \quad\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{20}}$ On comparing both side we get- $\frac{t}{20}=3$ $t=20 \times 3$ $t=60 \text { years }$
