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NUCLEAR PHYSICS

147835 A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

1

2 \log_{e} 2

2

0.5 \log_{e} 2

3

0.6 \log_{e} 2

4

0.8 \log_{e} 2

Explanation:

B Given that,
$\frac{dN}{dt} = 10000 per minuts$
$\frac{dN}{dt} = λ N_{0}$
$10000 = λ N_{0}$
After 4 minutes,
$2500 = λ N$
We know that, $N = N_{0} e^{- 4 λ}$
$N λ = N_{0} λ e^{- 4 λ}$
$2500 = 10000 e^{- 4 λ}$
$e^{- 4 λ} = \frac{1}{4}$
Taking natural $\log$ both side, we get-
$\ln e^{- 4 λ} = \ln (\frac{1}{4})$
$- 4 λ \ln = \ln (2^{- 2})$
$- 4 λ = - 2 \ln 2$
$λ = 0.5 \ln 2$
$λ = 0.5 \log_{e} 2$

MHT-CET 2017

NUCLEAR PHYSICS

147836 The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

1

\log_{e} 2 / 5

2

\frac{5}{\log_{e} 2}

3

5 \log 10^{2}

4

5 \log_{e} 2

Explanation:

D Given that,
$R_{o} = N_{o}$ count per minute
$R = \frac{N_{o}}{e}$ count per minute
$t = 5$ minute
From activity law, $R = R_{o} e^{- λ t}$
$\frac{N_{o}}{e} = N_{o} e^{- 5 λ}$
$e^{- 1} = e^{- 5 λ}$
$5 λ = 1$
$λ = 1 / 5 per minute$
At, $t = T_{1 / 2} =$ The activity $R$ reduces to $R_{0} / 2$
From equation (i), we get -
$\frac{R_{o}}{2} = R_{o} e^{- λ T_{1 / 2}}$
$2 = e^{- λ T_{1 / 2}}$
Logarithms of both sides, we get -
$λ T_{1 / 2} = \log_{e} 2$
$T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{\log_{e} 2}{1 / 5}$
$T_{1 / 2} = 5 \log_{e} 2 minutes [∵ λ = \frac{1}{5}]$

AIIMS-2017

NUCLEAR PHYSICS

147837 A radioactive element $A$ decay in stable element $B$ , initially a fresh sample of $A$ is available. In this sample variation in number of nuclei of $B$ with time is shown by

1 a

2 b

3 c

4 d

Explanation:

A We know that, the number of nuclide at time $t$ $-$
$N = N_{0} e^{- λ t}$
Where, $N_{0} =$ initial number of nuclide
This equation is equivalent, $y = a e^{- k x}$

Hence, the radioactive element A decay in stable element $B$ which shows exponential decay.

CG PET- 2008

NUCLEAR PHYSICS

147838 If half-life of a radioactive atom is 2.3 days, then its decay constant would be

1 0.1

2 0.2

3 0.3

4 2.3

Explanation:

C Given that,
Half-life of radioactive atom $(T_{1 / 2}) = 2.3$ day
$λ =$ ?
We know that,
Decay constant $(λ) = (\frac{0.693}{T_{1 / 2}})$
$λ = \frac{0.693}{2.3}$
$λ = \frac{6930}{23000}$
$λ = 0.3$

CG PET- 2007

NUCLEAR PHYSICS

147839 If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2} = 5 yrs$ , then the amount of substance left after $15 yr$ is

1

\frac{N_{0}}{8}

2

\frac{N_{0}}{16}

3

\frac{N_{0}}{2}

4

\frac{N_{0}}{4}

Explanation:

A Given that,
Original mass of substance $= N_{o}$
$T_{1 / 2} = 5$ year
Time $(t) = 15 yrs$
Amount left after decay is given by $N$ ,
$N = N_{0} {(\frac{1}{2})}^{n}$
Number of half-lives $(n) = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
Putting the value of $n$ , we get -
$N = N_{0} {(\frac{1}{2})}^{3}$
$N = \frac{N_{0}}{8}$

CG PET- 2006

NUCLEAR PHYSICS

147835 A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

1

2 \log_{e} 2

2

0.5 \log_{e} 2

3

0.6 \log_{e} 2

4

0.8 \log_{e} 2

Explanation:

B Given that,
$\frac{dN}{dt} = 10000 per minuts$
$\frac{dN}{dt} = λ N_{0}$
$10000 = λ N_{0}$
After 4 minutes,
$2500 = λ N$
We know that, $N = N_{0} e^{- 4 λ}$
$N λ = N_{0} λ e^{- 4 λ}$
$2500 = 10000 e^{- 4 λ}$
$e^{- 4 λ} = \frac{1}{4}$
Taking natural $\log$ both side, we get-
$\ln e^{- 4 λ} = \ln (\frac{1}{4})$
$- 4 λ \ln = \ln (2^{- 2})$
$- 4 λ = - 2 \ln 2$
$λ = 0.5 \ln 2$
$λ = 0.5 \log_{e} 2$

MHT-CET 2017

NUCLEAR PHYSICS

147836 The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

1

\log_{e} 2 / 5

2

\frac{5}{\log_{e} 2}

3

5 \log 10^{2}

4

5 \log_{e} 2

Explanation:

D Given that,
$R_{o} = N_{o}$ count per minute
$R = \frac{N_{o}}{e}$ count per minute
$t = 5$ minute
From activity law, $R = R_{o} e^{- λ t}$
$\frac{N_{o}}{e} = N_{o} e^{- 5 λ}$
$e^{- 1} = e^{- 5 λ}$
$5 λ = 1$
$λ = 1 / 5 per minute$
At, $t = T_{1 / 2} =$ The activity $R$ reduces to $R_{0} / 2$
From equation (i), we get -
$\frac{R_{o}}{2} = R_{o} e^{- λ T_{1 / 2}}$
$2 = e^{- λ T_{1 / 2}}$
Logarithms of both sides, we get -
$λ T_{1 / 2} = \log_{e} 2$
$T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{\log_{e} 2}{1 / 5}$
$T_{1 / 2} = 5 \log_{e} 2 minutes [∵ λ = \frac{1}{5}]$

AIIMS-2017

NUCLEAR PHYSICS

147837 A radioactive element $A$ decay in stable element $B$ , initially a fresh sample of $A$ is available. In this sample variation in number of nuclei of $B$ with time is shown by

1 a

2 b

3 c

4 d

Explanation:

A We know that, the number of nuclide at time $t$ $-$
$N = N_{0} e^{- λ t}$
Where, $N_{0} =$ initial number of nuclide
This equation is equivalent, $y = a e^{- k x}$

Hence, the radioactive element A decay in stable element $B$ which shows exponential decay.

CG PET- 2008

NUCLEAR PHYSICS

147838 If half-life of a radioactive atom is 2.3 days, then its decay constant would be

1 0.1

2 0.2

3 0.3

4 2.3

Explanation:

C Given that,
Half-life of radioactive atom $(T_{1 / 2}) = 2.3$ day
$λ =$ ?
We know that,
Decay constant $(λ) = (\frac{0.693}{T_{1 / 2}})$
$λ = \frac{0.693}{2.3}$
$λ = \frac{6930}{23000}$
$λ = 0.3$

CG PET- 2007

NUCLEAR PHYSICS

147839 If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2} = 5 yrs$ , then the amount of substance left after $15 yr$ is

1

\frac{N_{0}}{8}

2

\frac{N_{0}}{16}

3

\frac{N_{0}}{2}

4

\frac{N_{0}}{4}

Explanation:

A Given that,
Original mass of substance $= N_{o}$
$T_{1 / 2} = 5$ year
Time $(t) = 15 yrs$
Amount left after decay is given by $N$ ,
$N = N_{0} {(\frac{1}{2})}^{n}$
Number of half-lives $(n) = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
Putting the value of $n$ , we get -
$N = N_{0} {(\frac{1}{2})}^{3}$
$N = \frac{N_{0}}{8}$

CG PET- 2006

NUCLEAR PHYSICS

147835 A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

1

2 \log_{e} 2

2

0.5 \log_{e} 2

3

0.6 \log_{e} 2

4

0.8 \log_{e} 2

Explanation:

B Given that,
$\frac{dN}{dt} = 10000 per minuts$
$\frac{dN}{dt} = λ N_{0}$
$10000 = λ N_{0}$
After 4 minutes,
$2500 = λ N$
We know that, $N = N_{0} e^{- 4 λ}$
$N λ = N_{0} λ e^{- 4 λ}$
$2500 = 10000 e^{- 4 λ}$
$e^{- 4 λ} = \frac{1}{4}$
Taking natural $\log$ both side, we get-
$\ln e^{- 4 λ} = \ln (\frac{1}{4})$
$- 4 λ \ln = \ln (2^{- 2})$
$- 4 λ = - 2 \ln 2$
$λ = 0.5 \ln 2$
$λ = 0.5 \log_{e} 2$

MHT-CET 2017

NUCLEAR PHYSICS

147836 The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

1

\log_{e} 2 / 5

2

\frac{5}{\log_{e} 2}

3

5 \log 10^{2}

4

5 \log_{e} 2

Explanation:

D Given that,
$R_{o} = N_{o}$ count per minute
$R = \frac{N_{o}}{e}$ count per minute
$t = 5$ minute
From activity law, $R = R_{o} e^{- λ t}$
$\frac{N_{o}}{e} = N_{o} e^{- 5 λ}$
$e^{- 1} = e^{- 5 λ}$
$5 λ = 1$
$λ = 1 / 5 per minute$
At, $t = T_{1 / 2} =$ The activity $R$ reduces to $R_{0} / 2$
From equation (i), we get -
$\frac{R_{o}}{2} = R_{o} e^{- λ T_{1 / 2}}$
$2 = e^{- λ T_{1 / 2}}$
Logarithms of both sides, we get -
$λ T_{1 / 2} = \log_{e} 2$
$T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{\log_{e} 2}{1 / 5}$
$T_{1 / 2} = 5 \log_{e} 2 minutes [∵ λ = \frac{1}{5}]$

AIIMS-2017

NUCLEAR PHYSICS

147837 A radioactive element $A$ decay in stable element $B$ , initially a fresh sample of $A$ is available. In this sample variation in number of nuclei of $B$ with time is shown by

1 a

2 b

3 c

4 d

Explanation:

A We know that, the number of nuclide at time $t$ $-$
$N = N_{0} e^{- λ t}$
Where, $N_{0} =$ initial number of nuclide
This equation is equivalent, $y = a e^{- k x}$

Hence, the radioactive element A decay in stable element $B$ which shows exponential decay.

CG PET- 2008

NUCLEAR PHYSICS

147838 If half-life of a radioactive atom is 2.3 days, then its decay constant would be

1 0.1

2 0.2

3 0.3

4 2.3

Explanation:

C Given that,
Half-life of radioactive atom $(T_{1 / 2}) = 2.3$ day
$λ =$ ?
We know that,
Decay constant $(λ) = (\frac{0.693}{T_{1 / 2}})$
$λ = \frac{0.693}{2.3}$
$λ = \frac{6930}{23000}$
$λ = 0.3$

CG PET- 2007

NUCLEAR PHYSICS

147839 If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2} = 5 yrs$ , then the amount of substance left after $15 yr$ is

1

\frac{N_{0}}{8}

2

\frac{N_{0}}{16}

3

\frac{N_{0}}{2}

4

\frac{N_{0}}{4}

Explanation:

A Given that,
Original mass of substance $= N_{o}$
$T_{1 / 2} = 5$ year
Time $(t) = 15 yrs$
Amount left after decay is given by $N$ ,
$N = N_{0} {(\frac{1}{2})}^{n}$
Number of half-lives $(n) = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
Putting the value of $n$ , we get -
$N = N_{0} {(\frac{1}{2})}^{3}$
$N = \frac{N_{0}}{8}$

CG PET- 2006

NUCLEAR PHYSICS

147835 A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

1

2 \log_{e} 2

2

0.5 \log_{e} 2

3

0.6 \log_{e} 2

4

0.8 \log_{e} 2

Explanation:

B Given that,
$\frac{dN}{dt} = 10000 per minuts$
$\frac{dN}{dt} = λ N_{0}$
$10000 = λ N_{0}$
After 4 minutes,
$2500 = λ N$
We know that, $N = N_{0} e^{- 4 λ}$
$N λ = N_{0} λ e^{- 4 λ}$
$2500 = 10000 e^{- 4 λ}$
$e^{- 4 λ} = \frac{1}{4}$
Taking natural $\log$ both side, we get-
$\ln e^{- 4 λ} = \ln (\frac{1}{4})$
$- 4 λ \ln = \ln (2^{- 2})$
$- 4 λ = - 2 \ln 2$
$λ = 0.5 \ln 2$
$λ = 0.5 \log_{e} 2$

MHT-CET 2017

NUCLEAR PHYSICS

147836 The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

1

\log_{e} 2 / 5

2

\frac{5}{\log_{e} 2}

3

5 \log 10^{2}

4

5 \log_{e} 2

Explanation:

D Given that,
$R_{o} = N_{o}$ count per minute
$R = \frac{N_{o}}{e}$ count per minute
$t = 5$ minute
From activity law, $R = R_{o} e^{- λ t}$
$\frac{N_{o}}{e} = N_{o} e^{- 5 λ}$
$e^{- 1} = e^{- 5 λ}$
$5 λ = 1$
$λ = 1 / 5 per minute$
At, $t = T_{1 / 2} =$ The activity $R$ reduces to $R_{0} / 2$
From equation (i), we get -
$\frac{R_{o}}{2} = R_{o} e^{- λ T_{1 / 2}}$
$2 = e^{- λ T_{1 / 2}}$
Logarithms of both sides, we get -
$λ T_{1 / 2} = \log_{e} 2$
$T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{\log_{e} 2}{1 / 5}$
$T_{1 / 2} = 5 \log_{e} 2 minutes [∵ λ = \frac{1}{5}]$

AIIMS-2017

NUCLEAR PHYSICS

147837 A radioactive element $A$ decay in stable element $B$ , initially a fresh sample of $A$ is available. In this sample variation in number of nuclei of $B$ with time is shown by

1 a

2 b

3 c

4 d

Explanation:

A We know that, the number of nuclide at time $t$ $-$
$N = N_{0} e^{- λ t}$
Where, $N_{0} =$ initial number of nuclide
This equation is equivalent, $y = a e^{- k x}$

Hence, the radioactive element A decay in stable element $B$ which shows exponential decay.

CG PET- 2008

NUCLEAR PHYSICS

147838 If half-life of a radioactive atom is 2.3 days, then its decay constant would be

1 0.1

2 0.2

3 0.3

4 2.3

Explanation:

C Given that,
Half-life of radioactive atom $(T_{1 / 2}) = 2.3$ day
$λ =$ ?
We know that,
Decay constant $(λ) = (\frac{0.693}{T_{1 / 2}})$
$λ = \frac{0.693}{2.3}$
$λ = \frac{6930}{23000}$
$λ = 0.3$

CG PET- 2007

NUCLEAR PHYSICS

147839 If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2} = 5 yrs$ , then the amount of substance left after $15 yr$ is

1

\frac{N_{0}}{8}

2

\frac{N_{0}}{16}

3

\frac{N_{0}}{2}

4

\frac{N_{0}}{4}

Explanation:

A Given that,
Original mass of substance $= N_{o}$
$T_{1 / 2} = 5$ year
Time $(t) = 15 yrs$
Amount left after decay is given by $N$ ,
$N = N_{0} {(\frac{1}{2})}^{n}$
Number of half-lives $(n) = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
Putting the value of $n$ , we get -
$N = N_{0} {(\frac{1}{2})}^{3}$
$N = \frac{N_{0}}{8}$

CG PET- 2006

NUCLEAR PHYSICS

147835 A radioactive element has rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

1

2 \log_{e} 2

2

0.5 \log_{e} 2

3

0.6 \log_{e} 2

4

0.8 \log_{e} 2

Explanation:

B Given that,
$\frac{dN}{dt} = 10000 per minuts$
$\frac{dN}{dt} = λ N_{0}$
$10000 = λ N_{0}$
After 4 minutes,
$2500 = λ N$
We know that, $N = N_{0} e^{- 4 λ}$
$N λ = N_{0} λ e^{- 4 λ}$
$2500 = 10000 e^{- 4 λ}$
$e^{- 4 λ} = \frac{1}{4}$
Taking natural $\log$ both side, we get-
$\ln e^{- 4 λ} = \ln (\frac{1}{4})$
$- 4 λ \ln = \ln (2^{- 2})$
$- 4 λ = - 2 \ln 2$
$λ = 0.5 \ln 2$
$λ = 0.5 \log_{e} 2$

MHT-CET 2017

NUCLEAR PHYSICS

147836 The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t = 0$ and $N_{0} / e$ counts per minute at $t = 5$ minutes. The time (in minutes) at which the activity reduces to half its value is

1

\log_{e} 2 / 5

2

\frac{5}{\log_{e} 2}

3

5 \log 10^{2}

4

5 \log_{e} 2

Explanation:

D Given that,
$R_{o} = N_{o}$ count per minute
$R = \frac{N_{o}}{e}$ count per minute
$t = 5$ minute
From activity law, $R = R_{o} e^{- λ t}$
$\frac{N_{o}}{e} = N_{o} e^{- 5 λ}$
$e^{- 1} = e^{- 5 λ}$
$5 λ = 1$
$λ = 1 / 5 per minute$
At, $t = T_{1 / 2} =$ The activity $R$ reduces to $R_{0} / 2$
From equation (i), we get -
$\frac{R_{o}}{2} = R_{o} e^{- λ T_{1 / 2}}$
$2 = e^{- λ T_{1 / 2}}$
Logarithms of both sides, we get -
$λ T_{1 / 2} = \log_{e} 2$
$T_{1 / 2} = \frac{\log_{e} 2}{λ} = \frac{\log_{e} 2}{1 / 5}$
$T_{1 / 2} = 5 \log_{e} 2 minutes [∵ λ = \frac{1}{5}]$

AIIMS-2017

NUCLEAR PHYSICS

147837 A radioactive element $A$ decay in stable element $B$ , initially a fresh sample of $A$ is available. In this sample variation in number of nuclei of $B$ with time is shown by

1 a

2 b

3 c

4 d

Explanation:

A We know that, the number of nuclide at time $t$ $-$
$N = N_{0} e^{- λ t}$
Where, $N_{0} =$ initial number of nuclide
This equation is equivalent, $y = a e^{- k x}$

Hence, the radioactive element A decay in stable element $B$ which shows exponential decay.

CG PET- 2008

NUCLEAR PHYSICS

147838 If half-life of a radioactive atom is 2.3 days, then its decay constant would be

1 0.1

2 0.2

3 0.3

4 2.3

Explanation:

C Given that,
Half-life of radioactive atom $(T_{1 / 2}) = 2.3$ day
$λ =$ ?
We know that,
Decay constant $(λ) = (\frac{0.693}{T_{1 / 2}})$
$λ = \frac{0.693}{2.3}$
$λ = \frac{6930}{23000}$
$λ = 0.3$

CG PET- 2007

NUCLEAR PHYSICS

147839 If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2} = 5 yrs$ , then the amount of substance left after $15 yr$ is

1

\frac{N_{0}}{8}

2

\frac{N_{0}}{16}

3

\frac{N_{0}}{2}

4

\frac{N_{0}}{4}

Explanation:

A Given that,
Original mass of substance $= N_{o}$
$T_{1 / 2} = 5$ year
Time $(t) = 15 yrs$
Amount left after decay is given by $N$ ,
$N = N_{0} {(\frac{1}{2})}^{n}$
Number of half-lives $(n) = \frac{t}{T_{1 / 2}} = \frac{15}{5} = 3$
Putting the value of $n$ , we get -
$N = N_{0} {(\frac{1}{2})}^{3}$
$N = \frac{N_{0}}{8}$

CG PET- 2006