QBManager

NUCLEAR PHYSICS

147826 If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $75 %$ .

1 27.7 years

2 57.7 years

3 60 years

4 87 years

Explanation:

A Given that,
Decay constant $(λ) = 0.05 /$ year
When $75 %$ decay,
$N = N_{0} - \frac{3 N_{0}}{4}$
$N = \frac{N_{0}}{4}$
We know that,
$\ln \frac{N}{N_{0}} = - λ t$
$\ln (\frac{1}{4}) = - λ t$
$- 1.386 = - λ t$
$λ = \frac{1.386}{t} = \frac{1.386}{0.05}$
$t = 27.72 years$

AIIMS-27.05.2018(M)

NUCLEAR PHYSICS

147827 Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $50$ 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

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Radioactivity $= - \frac{dN}{dt} = λ N$
$∵ λ N = \frac{0.693 N}{T_{1 / 2}}$
$λ N = \frac{0.693 \times 10^{8}}{50} = \frac{0.693 \times 1.2 \times 10^{8}}{60}$
$λ N = 0.693 \times 2 \times 10^{6}$
Radioactive is inversely proportional to half-life period,
$∴ R \propto \frac{1}{T_{1 / 2}}$

AIIMS-27.05.2018(E)

NUCLEAR PHYSICS

147828 Radioactive material $A$ has decay constant $8 λ$ and material $B$ has decay constant $λ$ . Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$ ?

1

\frac{1}{λ}

2

\frac{1}{7 λ}

3

\frac{1}{8 λ}

4

\frac{1}{9 λ}

Explanation:

B Considering first order decay process and same number of initial number of atom.
We know that,
$So, N = N_{0} e^{- λ t}$
$and N_{1} = N_{0} e^{- 8 λ t}$
$N_{2} = N_{0} e^{- λ t}$
on dividing eqaution (i) by (ii) we get -
$\frac{N_{1}}{N_{2}} = \frac{N_{0} e^{- 8 λ t}}{N_{0} e^{- λ t}}$
$\frac{N_{1}}{N_{2}} = e^{- 7 λ t}$
Given that,
$\frac{N_{1}}{N_{2}} = \frac{1}{e}$
$Hence, \frac{1}{e} = e^{- 7 λ t}$
$e^{- 1} = e^{- 7 λ t}$
On comparing both side we get -
$- 7 λ t = - 1$
$t = \frac{1}{7 λ}$

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 $(T_{1 / 2}) = 20$ years According to question,
$\frac{N}{N_{o} - N} = \frac{1}{7}$
$7 N = N_{o} - N$
$8 N = N_{o}$
$\frac{N}{N_{o}} = \frac{1}{8}$
We know that, radioactive equation,
$N = N_{o} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{20}}$
On comparing both side we get-
$\frac{t}{20} = 3$
$t = 20 \times 3$
$t = 60 years$

AP EAMCET-28.04.2017

NUCLEAR PHYSICS

147826 If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $75 %$ .

1 27.7 years

2 57.7 years

3 60 years

4 87 years

Explanation:

A Given that,
Decay constant $(λ) = 0.05 /$ year
When $75 %$ decay,
$N = N_{0} - \frac{3 N_{0}}{4}$
$N = \frac{N_{0}}{4}$
We know that,
$\ln \frac{N}{N_{0}} = - λ t$
$\ln (\frac{1}{4}) = - λ t$
$- 1.386 = - λ t$
$λ = \frac{1.386}{t} = \frac{1.386}{0.05}$
$t = 27.72 years$

AIIMS-27.05.2018(M)

NUCLEAR PHYSICS

147827 Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $50$ 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

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Radioactivity $= - \frac{dN}{dt} = λ N$
$∵ λ N = \frac{0.693 N}{T_{1 / 2}}$
$λ N = \frac{0.693 \times 10^{8}}{50} = \frac{0.693 \times 1.2 \times 10^{8}}{60}$
$λ N = 0.693 \times 2 \times 10^{6}$
Radioactive is inversely proportional to half-life period,
$∴ R \propto \frac{1}{T_{1 / 2}}$

AIIMS-27.05.2018(E)

NUCLEAR PHYSICS

147828 Radioactive material $A$ has decay constant $8 λ$ and material $B$ has decay constant $λ$ . Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$ ?

1

\frac{1}{λ}

2

\frac{1}{7 λ}

3

\frac{1}{8 λ}

4

\frac{1}{9 λ}

Explanation:

B Considering first order decay process and same number of initial number of atom.
We know that,
$So, N = N_{0} e^{- λ t}$
$and N_{1} = N_{0} e^{- 8 λ t}$
$N_{2} = N_{0} e^{- λ t}$
on dividing eqaution (i) by (ii) we get -
$\frac{N_{1}}{N_{2}} = \frac{N_{0} e^{- 8 λ t}}{N_{0} e^{- λ t}}$
$\frac{N_{1}}{N_{2}} = e^{- 7 λ t}$
Given that,
$\frac{N_{1}}{N_{2}} = \frac{1}{e}$
$Hence, \frac{1}{e} = e^{- 7 λ t}$
$e^{- 1} = e^{- 7 λ t}$
On comparing both side we get -
$- 7 λ t = - 1$
$t = \frac{1}{7 λ}$

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 $(T_{1 / 2}) = 20$ years According to question,
$\frac{N}{N_{o} - N} = \frac{1}{7}$
$7 N = N_{o} - N$
$8 N = N_{o}$
$\frac{N}{N_{o}} = \frac{1}{8}$
We know that, radioactive equation,
$N = N_{o} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{20}}$
On comparing both side we get-
$\frac{t}{20} = 3$
$t = 20 \times 3$
$t = 60 years$

AP EAMCET-28.04.2017

NUCLEAR PHYSICS

147826 If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $75 %$ .

1 27.7 years

2 57.7 years

3 60 years

4 87 years

Explanation:

A Given that,
Decay constant $(λ) = 0.05 /$ year
When $75 %$ decay,
$N = N_{0} - \frac{3 N_{0}}{4}$
$N = \frac{N_{0}}{4}$
We know that,
$\ln \frac{N}{N_{0}} = - λ t$
$\ln (\frac{1}{4}) = - λ t$
$- 1.386 = - λ t$
$λ = \frac{1.386}{t} = \frac{1.386}{0.05}$
$t = 27.72 years$

AIIMS-27.05.2018(M)

NUCLEAR PHYSICS

147827 Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $50$ 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

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Radioactivity $= - \frac{dN}{dt} = λ N$
$∵ λ N = \frac{0.693 N}{T_{1 / 2}}$
$λ N = \frac{0.693 \times 10^{8}}{50} = \frac{0.693 \times 1.2 \times 10^{8}}{60}$
$λ N = 0.693 \times 2 \times 10^{6}$
Radioactive is inversely proportional to half-life period,
$∴ R \propto \frac{1}{T_{1 / 2}}$

AIIMS-27.05.2018(E)

NUCLEAR PHYSICS

147828 Radioactive material $A$ has decay constant $8 λ$ and material $B$ has decay constant $λ$ . Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$ ?

1

\frac{1}{λ}

2

\frac{1}{7 λ}

3

\frac{1}{8 λ}

4

\frac{1}{9 λ}

Explanation:

B Considering first order decay process and same number of initial number of atom.
We know that,
$So, N = N_{0} e^{- λ t}$
$and N_{1} = N_{0} e^{- 8 λ t}$
$N_{2} = N_{0} e^{- λ t}$
on dividing eqaution (i) by (ii) we get -
$\frac{N_{1}}{N_{2}} = \frac{N_{0} e^{- 8 λ t}}{N_{0} e^{- λ t}}$
$\frac{N_{1}}{N_{2}} = e^{- 7 λ t}$
Given that,
$\frac{N_{1}}{N_{2}} = \frac{1}{e}$
$Hence, \frac{1}{e} = e^{- 7 λ t}$
$e^{- 1} = e^{- 7 λ t}$
On comparing both side we get -
$- 7 λ t = - 1$
$t = \frac{1}{7 λ}$

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 $(T_{1 / 2}) = 20$ years According to question,
$\frac{N}{N_{o} - N} = \frac{1}{7}$
$7 N = N_{o} - N$
$8 N = N_{o}$
$\frac{N}{N_{o}} = \frac{1}{8}$
We know that, radioactive equation,
$N = N_{o} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{20}}$
On comparing both side we get-
$\frac{t}{20} = 3$
$t = 20 \times 3$
$t = 60 years$

AP EAMCET-28.04.2017

NUCLEAR PHYSICS

147826 If decay constant of a radioactive sample is $0.05 /$ year, then find out the time for which sample will decay by $75 %$ .

1 27.7 years

2 57.7 years

3 60 years

4 87 years

Explanation:

A Given that,
Decay constant $(λ) = 0.05 /$ year
When $75 %$ decay,
$N = N_{0} - \frac{3 N_{0}}{4}$
$N = \frac{N_{0}}{4}$
We know that,
$\ln \frac{N}{N_{0}} = - λ t$
$\ln (\frac{1}{4}) = - λ t$
$- 1.386 = - λ t$
$λ = \frac{1.386}{t} = \frac{1.386}{0.05}$
$t = 27.72 years$

AIIMS-27.05.2018(M)

NUCLEAR PHYSICS

147827 Assertion: Radioactivity of $10^{8}$ undecayed radioactive nuclei of half life of $50$ 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

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Radioactivity $= - \frac{dN}{dt} = λ N$
$∵ λ N = \frac{0.693 N}{T_{1 / 2}}$
$λ N = \frac{0.693 \times 10^{8}}{50} = \frac{0.693 \times 1.2 \times 10^{8}}{60}$
$λ N = 0.693 \times 2 \times 10^{6}$
Radioactive is inversely proportional to half-life period,
$∴ R \propto \frac{1}{T_{1 / 2}}$

AIIMS-27.05.2018(E)

NUCLEAR PHYSICS

147828 Radioactive material $A$ has decay constant $8 λ$ and material $B$ has decay constant $λ$ . Initially, they have same number of nuclei. After what time, the ratio of number of nuclei of material B to that A will be $\frac{1}{e}$ ?

1

\frac{1}{λ}

2

\frac{1}{7 λ}

3

\frac{1}{8 λ}

4

\frac{1}{9 λ}

Explanation:

B Considering first order decay process and same number of initial number of atom.
We know that,
$So, N = N_{0} e^{- λ t}$
$and N_{1} = N_{0} e^{- 8 λ t}$
$N_{2} = N_{0} e^{- λ t}$
on dividing eqaution (i) by (ii) we get -
$\frac{N_{1}}{N_{2}} = \frac{N_{0} e^{- 8 λ t}}{N_{0} e^{- λ t}}$
$\frac{N_{1}}{N_{2}} = e^{- 7 λ t}$
Given that,
$\frac{N_{1}}{N_{2}} = \frac{1}{e}$
$Hence, \frac{1}{e} = e^{- 7 λ t}$
$e^{- 1} = e^{- 7 λ t}$
On comparing both side we get -
$- 7 λ t = - 1$
$t = \frac{1}{7 λ}$

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 $(T_{1 / 2}) = 20$ years According to question,
$\frac{N}{N_{o} - N} = \frac{1}{7}$
$7 N = N_{o} - N$
$8 N = N_{o}$
$\frac{N}{N_{o}} = \frac{1}{8}$
We know that, radioactive equation,
$N = N_{o} {(\frac{1}{2})}^{n}$
$\frac{N}{N_{0}} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{1}{8} = {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{20}}$
On comparing both side we get-
$\frac{t}{20} = 3$
$t = 20 \times 3$
$t = 60 years$

AP EAMCET-28.04.2017