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NUCLEAR PHYSICS

147905 A radioactive element $x$ converts into another stable element. Half-life of $x$ is $2 h$ , initially only $x$ is present. After time t, the ratio of atoms of $x$ and $y$ is found to be $1 : 4$ , then $t$ in hour is

1 2

2 4

3 between 4 and 6

4 6

Explanation:

C $X \underset{T_{1 / 2} = 2 hrs}{\overset{λ}{⟶}} Y$
Number of atoms that decay in $t$ hours $= N_{Y}$
Number of that remain undecayed in $t$ hours $= N_{X}$
$∴ N = N_{o}, at t = 0$
$∴ N_{X} = N_{o} e^{- λ t}$
$N_{Y} = N_{o} - N_{X} = N_{o} (1 - e^{- λ t})$
$\frac{N_{X}}{N_{Y}} = \frac{e^{- λ t}}{1 - e^{- λ t}} = \frac{1}{4}$
$4 e^{- λ t} = 1 - e^{- λ t}$
$5 e^{- λ t} = 1$
$λ t = \ln 5$
$∵ t = \frac{1}{λ} \ln 5$
$or t = \frac{1}{\frac{\ln 2}{T_{1 / 2}}} \times \ln 5 = T_{1 / 2} \times \frac{\ln 5}{\ln 2} = 2 \frac{\ln 5}{\ln 2} hours$
$t = 4.644 hours$

UP CPMT-2009

NUCLEAR PHYSICS

147906 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays in 10 days will be

1

\frac{77}{80}

2

\frac{71}{80}

3

\frac{31}{32}

4

\frac{15}{16}

Explanation:

C We know that, radioactive law,
$N = N_{o} e^{- λ t}$
According to question,
$N = N_{o} - N_{o} \times \frac{7}{8} = \frac{N_{o}}{8}$
On solving equation (i), we get-
$\frac{N}{N_{o}} = e^{- λ t}$
$Or \frac{N_{o}}{N} = e^{λ t}$
$Taking natural log on both the side,$
$t = \frac{\ln (\frac{N_{o}}{N})}{λ}$
$t \propto \ln (\frac{N_{o}}{N})$
$So, \frac{t_{1}}{t_{2}} = \frac{\ln {(\frac{N_{o}}{N})}_{1}}{\ln {(\frac{N_{o}}{N})}_{2}}$
$\frac{6}{10} = \frac{\ln {(\frac{N_{o}}{N_{o} / 8})}_{1}}{\ln {(\frac{N_{o}}{N_{2}})}_{2}}$
$\frac{3}{5} = \frac{\ln 8}{\ln (\frac{N_{o}}{N_{2}})}$
$\ln (\frac{N_{o}}{N_{2}}) = \frac{5}{3} \ln 8$
$\ln (\frac{N_{o}}{N_{2}}) = \ln (8)^{5 / 3} = \ln 32$
$\frac{N_{o}}{N_{2}} = 32$
$or \frac{N_{2}}{N_{o}} = \frac{1}{32}$
Decay fraction $= N_{o} - \frac{N_{o}}{32} = \frac{31}{32} N_{o}$

UP CPMT-2014

NUCLEAR PHYSICS

147909 In a radioactive substance at $t = 0$ , the number of atoms is $8 \times 10^{4}$ , its half-life period is $3 yr$ . The number of atoms $1 \times 10^{4}$ will remain after interval

1

9 yr

2

8 yr

3

6 yr

4

24 yr

Explanation:

A Given that,
Initial number of radioactive substance $(N_{o}) = 8 \times 10^{4}$ Final number of radioactive substance $(N) = 1 \times 10^{4}$ Half-life of substance $(T_{1 / 2}) = 3$ yrs.
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$10^{4} = 8 \times 10^{4} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{10^{4}}{8 \times 10^{4}} = {(\frac{1}{2})}^{\frac{t}{3}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{3}}$
On comparing both the side, we get-
$\frac{t}{3} = 3$
$t = 9 y r$

UP CPMT-2010

NUCLEAR PHYSICS

147910 If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after 2 days, will be

1

\sqrt{2}

2

\frac{1}{\sqrt{2}}

3

\frac{\sqrt{2} - 1}{\sqrt{2}}

4

\frac{1}{2}

Explanation:

B Given that,
Half-life radioactive substance $(T_{1 / 2}) = 4$ days
Decay constant $(λ) = \frac{\ln 2}{T_{1 / 2}} = \frac{\ln 2}{4}$
Law of radioactivity,
$N = N_{o} e^{- λ t}$
After $t = 2$ day,
$N = N_{o} e^{- \frac{\ln 2}{4} \times 2}$
$N = N_{o} e^{- \frac{1}{2} \ln 2}$
$N = N_{o} e^{\ln 2^{- 1 / 2}}$
$(∵_{a} \ln b =_{b} \ln a)$
$N = N_{o} \times 2^{- 1 / 2}$
$N = \frac{N_{o}}{\sqrt{2}}$
$\frac{N}{N_{o}} = \frac{1}{\sqrt{2}}$

UP CPMT-2005

NUCLEAR PHYSICS

147905 A radioactive element $x$ converts into another stable element. Half-life of $x$ is $2 h$ , initially only $x$ is present. After time t, the ratio of atoms of $x$ and $y$ is found to be $1 : 4$ , then $t$ in hour is

1 2

2 4

3 between 4 and 6

4 6

Explanation:

C $X \underset{T_{1 / 2} = 2 hrs}{\overset{λ}{⟶}} Y$
Number of atoms that decay in $t$ hours $= N_{Y}$
Number of that remain undecayed in $t$ hours $= N_{X}$
$∴ N = N_{o}, at t = 0$
$∴ N_{X} = N_{o} e^{- λ t}$
$N_{Y} = N_{o} - N_{X} = N_{o} (1 - e^{- λ t})$
$\frac{N_{X}}{N_{Y}} = \frac{e^{- λ t}}{1 - e^{- λ t}} = \frac{1}{4}$
$4 e^{- λ t} = 1 - e^{- λ t}$
$5 e^{- λ t} = 1$
$λ t = \ln 5$
$∵ t = \frac{1}{λ} \ln 5$
$or t = \frac{1}{\frac{\ln 2}{T_{1 / 2}}} \times \ln 5 = T_{1 / 2} \times \frac{\ln 5}{\ln 2} = 2 \frac{\ln 5}{\ln 2} hours$
$t = 4.644 hours$

UP CPMT-2009

NUCLEAR PHYSICS

147906 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays in 10 days will be

1

\frac{77}{80}

2

\frac{71}{80}

3

\frac{31}{32}

4

\frac{15}{16}

Explanation:

C We know that, radioactive law,
$N = N_{o} e^{- λ t}$
According to question,
$N = N_{o} - N_{o} \times \frac{7}{8} = \frac{N_{o}}{8}$
On solving equation (i), we get-
$\frac{N}{N_{o}} = e^{- λ t}$
$Or \frac{N_{o}}{N} = e^{λ t}$
$Taking natural log on both the side,$
$t = \frac{\ln (\frac{N_{o}}{N})}{λ}$
$t \propto \ln (\frac{N_{o}}{N})$
$So, \frac{t_{1}}{t_{2}} = \frac{\ln {(\frac{N_{o}}{N})}_{1}}{\ln {(\frac{N_{o}}{N})}_{2}}$
$\frac{6}{10} = \frac{\ln {(\frac{N_{o}}{N_{o} / 8})}_{1}}{\ln {(\frac{N_{o}}{N_{2}})}_{2}}$
$\frac{3}{5} = \frac{\ln 8}{\ln (\frac{N_{o}}{N_{2}})}$
$\ln (\frac{N_{o}}{N_{2}}) = \frac{5}{3} \ln 8$
$\ln (\frac{N_{o}}{N_{2}}) = \ln (8)^{5 / 3} = \ln 32$
$\frac{N_{o}}{N_{2}} = 32$
$or \frac{N_{2}}{N_{o}} = \frac{1}{32}$
Decay fraction $= N_{o} - \frac{N_{o}}{32} = \frac{31}{32} N_{o}$

UP CPMT-2014

NUCLEAR PHYSICS

147909 In a radioactive substance at $t = 0$ , the number of atoms is $8 \times 10^{4}$ , its half-life period is $3 yr$ . The number of atoms $1 \times 10^{4}$ will remain after interval

1

9 yr

2

8 yr

3

6 yr

4

24 yr

Explanation:

A Given that,
Initial number of radioactive substance $(N_{o}) = 8 \times 10^{4}$ Final number of radioactive substance $(N) = 1 \times 10^{4}$ Half-life of substance $(T_{1 / 2}) = 3$ yrs.
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$10^{4} = 8 \times 10^{4} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{10^{4}}{8 \times 10^{4}} = {(\frac{1}{2})}^{\frac{t}{3}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{3}}$
On comparing both the side, we get-
$\frac{t}{3} = 3$
$t = 9 y r$

UP CPMT-2010

NUCLEAR PHYSICS

147910 If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after 2 days, will be

1

\sqrt{2}

2

\frac{1}{\sqrt{2}}

3

\frac{\sqrt{2} - 1}{\sqrt{2}}

4

\frac{1}{2}

Explanation:

B Given that,
Half-life radioactive substance $(T_{1 / 2}) = 4$ days
Decay constant $(λ) = \frac{\ln 2}{T_{1 / 2}} = \frac{\ln 2}{4}$
Law of radioactivity,
$N = N_{o} e^{- λ t}$
After $t = 2$ day,
$N = N_{o} e^{- \frac{\ln 2}{4} \times 2}$
$N = N_{o} e^{- \frac{1}{2} \ln 2}$
$N = N_{o} e^{\ln 2^{- 1 / 2}}$
$(∵_{a} \ln b =_{b} \ln a)$
$N = N_{o} \times 2^{- 1 / 2}$
$N = \frac{N_{o}}{\sqrt{2}}$
$\frac{N}{N_{o}} = \frac{1}{\sqrt{2}}$

UP CPMT-2005

NUCLEAR PHYSICS

147905 A radioactive element $x$ converts into another stable element. Half-life of $x$ is $2 h$ , initially only $x$ is present. After time t, the ratio of atoms of $x$ and $y$ is found to be $1 : 4$ , then $t$ in hour is

1 2

2 4

3 between 4 and 6

4 6

Explanation:

C $X \underset{T_{1 / 2} = 2 hrs}{\overset{λ}{⟶}} Y$
Number of atoms that decay in $t$ hours $= N_{Y}$
Number of that remain undecayed in $t$ hours $= N_{X}$
$∴ N = N_{o}, at t = 0$
$∴ N_{X} = N_{o} e^{- λ t}$
$N_{Y} = N_{o} - N_{X} = N_{o} (1 - e^{- λ t})$
$\frac{N_{X}}{N_{Y}} = \frac{e^{- λ t}}{1 - e^{- λ t}} = \frac{1}{4}$
$4 e^{- λ t} = 1 - e^{- λ t}$
$5 e^{- λ t} = 1$
$λ t = \ln 5$
$∵ t = \frac{1}{λ} \ln 5$
$or t = \frac{1}{\frac{\ln 2}{T_{1 / 2}}} \times \ln 5 = T_{1 / 2} \times \frac{\ln 5}{\ln 2} = 2 \frac{\ln 5}{\ln 2} hours$
$t = 4.644 hours$

UP CPMT-2009

NUCLEAR PHYSICS

147906 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays in 10 days will be

1

\frac{77}{80}

2

\frac{71}{80}

3

\frac{31}{32}

4

\frac{15}{16}

Explanation:

C We know that, radioactive law,
$N = N_{o} e^{- λ t}$
According to question,
$N = N_{o} - N_{o} \times \frac{7}{8} = \frac{N_{o}}{8}$
On solving equation (i), we get-
$\frac{N}{N_{o}} = e^{- λ t}$
$Or \frac{N_{o}}{N} = e^{λ t}$
$Taking natural log on both the side,$
$t = \frac{\ln (\frac{N_{o}}{N})}{λ}$
$t \propto \ln (\frac{N_{o}}{N})$
$So, \frac{t_{1}}{t_{2}} = \frac{\ln {(\frac{N_{o}}{N})}_{1}}{\ln {(\frac{N_{o}}{N})}_{2}}$
$\frac{6}{10} = \frac{\ln {(\frac{N_{o}}{N_{o} / 8})}_{1}}{\ln {(\frac{N_{o}}{N_{2}})}_{2}}$
$\frac{3}{5} = \frac{\ln 8}{\ln (\frac{N_{o}}{N_{2}})}$
$\ln (\frac{N_{o}}{N_{2}}) = \frac{5}{3} \ln 8$
$\ln (\frac{N_{o}}{N_{2}}) = \ln (8)^{5 / 3} = \ln 32$
$\frac{N_{o}}{N_{2}} = 32$
$or \frac{N_{2}}{N_{o}} = \frac{1}{32}$
Decay fraction $= N_{o} - \frac{N_{o}}{32} = \frac{31}{32} N_{o}$

UP CPMT-2014

NUCLEAR PHYSICS

147909 In a radioactive substance at $t = 0$ , the number of atoms is $8 \times 10^{4}$ , its half-life period is $3 yr$ . The number of atoms $1 \times 10^{4}$ will remain after interval

1

9 yr

2

8 yr

3

6 yr

4

24 yr

Explanation:

A Given that,
Initial number of radioactive substance $(N_{o}) = 8 \times 10^{4}$ Final number of radioactive substance $(N) = 1 \times 10^{4}$ Half-life of substance $(T_{1 / 2}) = 3$ yrs.
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$10^{4} = 8 \times 10^{4} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{10^{4}}{8 \times 10^{4}} = {(\frac{1}{2})}^{\frac{t}{3}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{3}}$
On comparing both the side, we get-
$\frac{t}{3} = 3$
$t = 9 y r$

UP CPMT-2010

NUCLEAR PHYSICS

147910 If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after 2 days, will be

1

\sqrt{2}

2

\frac{1}{\sqrt{2}}

3

\frac{\sqrt{2} - 1}{\sqrt{2}}

4

\frac{1}{2}

Explanation:

B Given that,
Half-life radioactive substance $(T_{1 / 2}) = 4$ days
Decay constant $(λ) = \frac{\ln 2}{T_{1 / 2}} = \frac{\ln 2}{4}$
Law of radioactivity,
$N = N_{o} e^{- λ t}$
After $t = 2$ day,
$N = N_{o} e^{- \frac{\ln 2}{4} \times 2}$
$N = N_{o} e^{- \frac{1}{2} \ln 2}$
$N = N_{o} e^{\ln 2^{- 1 / 2}}$
$(∵_{a} \ln b =_{b} \ln a)$
$N = N_{o} \times 2^{- 1 / 2}$
$N = \frac{N_{o}}{\sqrt{2}}$
$\frac{N}{N_{o}} = \frac{1}{\sqrt{2}}$

UP CPMT-2005

NUCLEAR PHYSICS

147905 A radioactive element $x$ converts into another stable element. Half-life of $x$ is $2 h$ , initially only $x$ is present. After time t, the ratio of atoms of $x$ and $y$ is found to be $1 : 4$ , then $t$ in hour is

1 2

2 4

3 between 4 and 6

4 6

Explanation:

C $X \underset{T_{1 / 2} = 2 hrs}{\overset{λ}{⟶}} Y$
Number of atoms that decay in $t$ hours $= N_{Y}$
Number of that remain undecayed in $t$ hours $= N_{X}$
$∴ N = N_{o}, at t = 0$
$∴ N_{X} = N_{o} e^{- λ t}$
$N_{Y} = N_{o} - N_{X} = N_{o} (1 - e^{- λ t})$
$\frac{N_{X}}{N_{Y}} = \frac{e^{- λ t}}{1 - e^{- λ t}} = \frac{1}{4}$
$4 e^{- λ t} = 1 - e^{- λ t}$
$5 e^{- λ t} = 1$
$λ t = \ln 5$
$∵ t = \frac{1}{λ} \ln 5$
$or t = \frac{1}{\frac{\ln 2}{T_{1 / 2}}} \times \ln 5 = T_{1 / 2} \times \frac{\ln 5}{\ln 2} = 2 \frac{\ln 5}{\ln 2} hours$
$t = 4.644 hours$

UP CPMT-2009

NUCLEAR PHYSICS

147906 The fraction of atoms of radioactive element that decays in 6 days is $7 / 8$ . The fraction that decays in 10 days will be

1

\frac{77}{80}

2

\frac{71}{80}

3

\frac{31}{32}

4

\frac{15}{16}

Explanation:

C We know that, radioactive law,
$N = N_{o} e^{- λ t}$
According to question,
$N = N_{o} - N_{o} \times \frac{7}{8} = \frac{N_{o}}{8}$
On solving equation (i), we get-
$\frac{N}{N_{o}} = e^{- λ t}$
$Or \frac{N_{o}}{N} = e^{λ t}$
$Taking natural log on both the side,$
$t = \frac{\ln (\frac{N_{o}}{N})}{λ}$
$t \propto \ln (\frac{N_{o}}{N})$
$So, \frac{t_{1}}{t_{2}} = \frac{\ln {(\frac{N_{o}}{N})}_{1}}{\ln {(\frac{N_{o}}{N})}_{2}}$
$\frac{6}{10} = \frac{\ln {(\frac{N_{o}}{N_{o} / 8})}_{1}}{\ln {(\frac{N_{o}}{N_{2}})}_{2}}$
$\frac{3}{5} = \frac{\ln 8}{\ln (\frac{N_{o}}{N_{2}})}$
$\ln (\frac{N_{o}}{N_{2}}) = \frac{5}{3} \ln 8$
$\ln (\frac{N_{o}}{N_{2}}) = \ln (8)^{5 / 3} = \ln 32$
$\frac{N_{o}}{N_{2}} = 32$
$or \frac{N_{2}}{N_{o}} = \frac{1}{32}$
Decay fraction $= N_{o} - \frac{N_{o}}{32} = \frac{31}{32} N_{o}$

UP CPMT-2014

NUCLEAR PHYSICS

147909 In a radioactive substance at $t = 0$ , the number of atoms is $8 \times 10^{4}$ , its half-life period is $3 yr$ . The number of atoms $1 \times 10^{4}$ will remain after interval

1

9 yr

2

8 yr

3

6 yr

4

24 yr

Explanation:

A Given that,
Initial number of radioactive substance $(N_{o}) = 8 \times 10^{4}$ Final number of radioactive substance $(N) = 1 \times 10^{4}$ Half-life of substance $(T_{1 / 2}) = 3$ yrs.
We know that, radioactive equation,
$N = N_{0} {(\frac{1}{2})}^{n}$
$10^{4} = 8 \times 10^{4} {(\frac{1}{2})}^{\frac{t}{T_{1 / 2}}} (∵ n = \frac{t}{T_{1 / 2}})$
$\frac{10^{4}}{8 \times 10^{4}} = {(\frac{1}{2})}^{\frac{t}{3}}$
${(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{\frac{t}{3}}$
On comparing both the side, we get-
$\frac{t}{3} = 3$
$t = 9 y r$

UP CPMT-2010

NUCLEAR PHYSICS

147910 If the half-life of any sample of radioactive substance is $4$ days, then the fraction of sample will remain undecayed after 2 days, will be

1

\sqrt{2}

2

\frac{1}{\sqrt{2}}

3

\frac{\sqrt{2} - 1}{\sqrt{2}}

4

\frac{1}{2}

Explanation:

B Given that,
Half-life radioactive substance $(T_{1 / 2}) = 4$ days
Decay constant $(λ) = \frac{\ln 2}{T_{1 / 2}} = \frac{\ln 2}{4}$
Law of radioactivity,
$N = N_{o} e^{- λ t}$
After $t = 2$ day,
$N = N_{o} e^{- \frac{\ln 2}{4} \times 2}$
$N = N_{o} e^{- \frac{1}{2} \ln 2}$
$N = N_{o} e^{\ln 2^{- 1 / 2}}$
$(∵_{a} \ln b =_{b} \ln a)$
$N = N_{o} \times 2^{- 1 / 2}$
$N = \frac{N_{o}}{\sqrt{2}}$
$\frac{N}{N_{o}} = \frac{1}{\sqrt{2}}$

UP CPMT-2005