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 $2 \log _{\mathrm{e}} 2$
2 $0.5 \log _{\mathrm{e}} 2$
3 $0.6 \log _{\mathrm{e}} 2$
4 $0.8 \log _{\mathrm{e}} 2$
Explanation:
B Given that, $\frac{\mathrm{dN}}{\mathrm{dt}}=10000 \text { per minuts }$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{0}$ $10000=\lambda \mathrm{N}_{0}$ After 4 minutes, $2500=\lambda \mathrm{N}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-4 \lambda}$ $\mathrm{N} \lambda=\mathrm{N}_{0} \lambda \mathrm{e}^{-4 \lambda}$ $2500=10000 \mathrm{e}^{-4 \lambda}$ $\mathrm{e}^{-4 \lambda}=\frac{1}{4}$ Taking natural $\log$ both side, we get- $\ln \mathrm{e}^{-4 \lambda}=\ln \left(\frac{1}{4}\right)$ $-4 \lambda \ln =\ln \left(2^{-2}\right)$ $-4 \lambda=-2 \ln 2$ $\lambda=0.5 \ln 2$ $\lambda=0.5 \log _{\mathrm{e}} 2$
MHT-CET 2017
NUCLEAR PHYSICS
147836
The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t=0$ and $\mathrm{N}_{0} / \mathrm{e}$ counts per minute at $\mathrm{t}=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
1 $\log _{\mathrm{e}} 2 / 5$
2 $\frac{5}{\log _{\mathrm{e}} 2}$
3 $5 \log 10^{2}$
4 $5 \log _{\mathrm{e}} 2$
Explanation:
D Given that, $\mathrm{R}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}}$ count per minute $\mathrm{R}=\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}$ count per minute $\mathrm{t}=5$ minute From activity law, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-5 \lambda}$ $\mathrm{e}^{-1}=\mathrm{e}^{-5 \lambda}$ $5 \lambda=1$ $\lambda=1 / 5 \text { per minute }$ At, $t=T_{1 / 2}=$ The activity $R$ reduces to $R_{0} / 2$ From equation (i), we get - $\frac{\mathrm{R}_{\mathrm{o}}}{2}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $2=\mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ Logarithms of both sides, we get - $\lambda \mathrm{T}_{1 / 2}=\log _{\mathrm{e}} 2$ $\mathrm{~T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}=\frac{\log _{\mathrm{e}} 2}{1 / 5}$ $\mathrm{~T}_{1 / 2}=5 \log _{\mathrm{e}} 2 \text { minutes } \quad\left[\because \lambda=\frac{1}{5}\right]$
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 $t$ $-$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Where, $\mathrm{N}_{0}=$ initial number of nuclide This equation is equivalent, $y=a e^{-k x}$ Hence, the radioactive element A decay in stable element $\mathrm{B}$ 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 $\left(\mathrm{T}_{1 / 2}\right)=2.3$ day $\lambda=$ ? We know that, Decay constant $(\lambda)=\left(\frac{0.693}{T_{1 / 2}}\right)$ $\lambda=\frac{0.693}{2.3}$ $\lambda=\frac{6930}{23000}$ $\lambda=0.3$
CG PET- 2007
NUCLEAR PHYSICS
147839
If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2}=5 \mathrm{yrs}$, then the amount of substance left after $15 \mathrm{yr}$ is
1 $\frac{\mathrm{N}_{0}}{8}$
2 $\frac{\mathrm{N}_{0}}{16}$
3 $\frac{\mathrm{N}_{0}}{2}$
4 $\frac{\mathrm{N}_{0}}{4}$
Explanation:
A Given that, Original mass of substance $=\mathrm{N}_{\mathrm{o}}$ $\mathrm{T}_{1 / 2}=5$ year Time $(\mathrm{t})=15 \mathrm{yrs}$ Amount left after decay is given by $\mathrm{N}$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Number of half-lives $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ Putting the value of $n$, we get - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$
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 $2 \log _{\mathrm{e}} 2$
2 $0.5 \log _{\mathrm{e}} 2$
3 $0.6 \log _{\mathrm{e}} 2$
4 $0.8 \log _{\mathrm{e}} 2$
Explanation:
B Given that, $\frac{\mathrm{dN}}{\mathrm{dt}}=10000 \text { per minuts }$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{0}$ $10000=\lambda \mathrm{N}_{0}$ After 4 minutes, $2500=\lambda \mathrm{N}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-4 \lambda}$ $\mathrm{N} \lambda=\mathrm{N}_{0} \lambda \mathrm{e}^{-4 \lambda}$ $2500=10000 \mathrm{e}^{-4 \lambda}$ $\mathrm{e}^{-4 \lambda}=\frac{1}{4}$ Taking natural $\log$ both side, we get- $\ln \mathrm{e}^{-4 \lambda}=\ln \left(\frac{1}{4}\right)$ $-4 \lambda \ln =\ln \left(2^{-2}\right)$ $-4 \lambda=-2 \ln 2$ $\lambda=0.5 \ln 2$ $\lambda=0.5 \log _{\mathrm{e}} 2$
MHT-CET 2017
NUCLEAR PHYSICS
147836
The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t=0$ and $\mathrm{N}_{0} / \mathrm{e}$ counts per minute at $\mathrm{t}=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
1 $\log _{\mathrm{e}} 2 / 5$
2 $\frac{5}{\log _{\mathrm{e}} 2}$
3 $5 \log 10^{2}$
4 $5 \log _{\mathrm{e}} 2$
Explanation:
D Given that, $\mathrm{R}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}}$ count per minute $\mathrm{R}=\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}$ count per minute $\mathrm{t}=5$ minute From activity law, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-5 \lambda}$ $\mathrm{e}^{-1}=\mathrm{e}^{-5 \lambda}$ $5 \lambda=1$ $\lambda=1 / 5 \text { per minute }$ At, $t=T_{1 / 2}=$ The activity $R$ reduces to $R_{0} / 2$ From equation (i), we get - $\frac{\mathrm{R}_{\mathrm{o}}}{2}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $2=\mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ Logarithms of both sides, we get - $\lambda \mathrm{T}_{1 / 2}=\log _{\mathrm{e}} 2$ $\mathrm{~T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}=\frac{\log _{\mathrm{e}} 2}{1 / 5}$ $\mathrm{~T}_{1 / 2}=5 \log _{\mathrm{e}} 2 \text { minutes } \quad\left[\because \lambda=\frac{1}{5}\right]$
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 $t$ $-$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Where, $\mathrm{N}_{0}=$ initial number of nuclide This equation is equivalent, $y=a e^{-k x}$ Hence, the radioactive element A decay in stable element $\mathrm{B}$ 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 $\left(\mathrm{T}_{1 / 2}\right)=2.3$ day $\lambda=$ ? We know that, Decay constant $(\lambda)=\left(\frac{0.693}{T_{1 / 2}}\right)$ $\lambda=\frac{0.693}{2.3}$ $\lambda=\frac{6930}{23000}$ $\lambda=0.3$
CG PET- 2007
NUCLEAR PHYSICS
147839
If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2}=5 \mathrm{yrs}$, then the amount of substance left after $15 \mathrm{yr}$ is
1 $\frac{\mathrm{N}_{0}}{8}$
2 $\frac{\mathrm{N}_{0}}{16}$
3 $\frac{\mathrm{N}_{0}}{2}$
4 $\frac{\mathrm{N}_{0}}{4}$
Explanation:
A Given that, Original mass of substance $=\mathrm{N}_{\mathrm{o}}$ $\mathrm{T}_{1 / 2}=5$ year Time $(\mathrm{t})=15 \mathrm{yrs}$ Amount left after decay is given by $\mathrm{N}$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Number of half-lives $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ Putting the value of $n$, we get - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$
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 $2 \log _{\mathrm{e}} 2$
2 $0.5 \log _{\mathrm{e}} 2$
3 $0.6 \log _{\mathrm{e}} 2$
4 $0.8 \log _{\mathrm{e}} 2$
Explanation:
B Given that, $\frac{\mathrm{dN}}{\mathrm{dt}}=10000 \text { per minuts }$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{0}$ $10000=\lambda \mathrm{N}_{0}$ After 4 minutes, $2500=\lambda \mathrm{N}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-4 \lambda}$ $\mathrm{N} \lambda=\mathrm{N}_{0} \lambda \mathrm{e}^{-4 \lambda}$ $2500=10000 \mathrm{e}^{-4 \lambda}$ $\mathrm{e}^{-4 \lambda}=\frac{1}{4}$ Taking natural $\log$ both side, we get- $\ln \mathrm{e}^{-4 \lambda}=\ln \left(\frac{1}{4}\right)$ $-4 \lambda \ln =\ln \left(2^{-2}\right)$ $-4 \lambda=-2 \ln 2$ $\lambda=0.5 \ln 2$ $\lambda=0.5 \log _{\mathrm{e}} 2$
MHT-CET 2017
NUCLEAR PHYSICS
147836
The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t=0$ and $\mathrm{N}_{0} / \mathrm{e}$ counts per minute at $\mathrm{t}=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
1 $\log _{\mathrm{e}} 2 / 5$
2 $\frac{5}{\log _{\mathrm{e}} 2}$
3 $5 \log 10^{2}$
4 $5 \log _{\mathrm{e}} 2$
Explanation:
D Given that, $\mathrm{R}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}}$ count per minute $\mathrm{R}=\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}$ count per minute $\mathrm{t}=5$ minute From activity law, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-5 \lambda}$ $\mathrm{e}^{-1}=\mathrm{e}^{-5 \lambda}$ $5 \lambda=1$ $\lambda=1 / 5 \text { per minute }$ At, $t=T_{1 / 2}=$ The activity $R$ reduces to $R_{0} / 2$ From equation (i), we get - $\frac{\mathrm{R}_{\mathrm{o}}}{2}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $2=\mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ Logarithms of both sides, we get - $\lambda \mathrm{T}_{1 / 2}=\log _{\mathrm{e}} 2$ $\mathrm{~T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}=\frac{\log _{\mathrm{e}} 2}{1 / 5}$ $\mathrm{~T}_{1 / 2}=5 \log _{\mathrm{e}} 2 \text { minutes } \quad\left[\because \lambda=\frac{1}{5}\right]$
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 $t$ $-$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Where, $\mathrm{N}_{0}=$ initial number of nuclide This equation is equivalent, $y=a e^{-k x}$ Hence, the radioactive element A decay in stable element $\mathrm{B}$ 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 $\left(\mathrm{T}_{1 / 2}\right)=2.3$ day $\lambda=$ ? We know that, Decay constant $(\lambda)=\left(\frac{0.693}{T_{1 / 2}}\right)$ $\lambda=\frac{0.693}{2.3}$ $\lambda=\frac{6930}{23000}$ $\lambda=0.3$
CG PET- 2007
NUCLEAR PHYSICS
147839
If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2}=5 \mathrm{yrs}$, then the amount of substance left after $15 \mathrm{yr}$ is
1 $\frac{\mathrm{N}_{0}}{8}$
2 $\frac{\mathrm{N}_{0}}{16}$
3 $\frac{\mathrm{N}_{0}}{2}$
4 $\frac{\mathrm{N}_{0}}{4}$
Explanation:
A Given that, Original mass of substance $=\mathrm{N}_{\mathrm{o}}$ $\mathrm{T}_{1 / 2}=5$ year Time $(\mathrm{t})=15 \mathrm{yrs}$ Amount left after decay is given by $\mathrm{N}$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Number of half-lives $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ Putting the value of $n$, we get - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$
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 $2 \log _{\mathrm{e}} 2$
2 $0.5 \log _{\mathrm{e}} 2$
3 $0.6 \log _{\mathrm{e}} 2$
4 $0.8 \log _{\mathrm{e}} 2$
Explanation:
B Given that, $\frac{\mathrm{dN}}{\mathrm{dt}}=10000 \text { per minuts }$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{0}$ $10000=\lambda \mathrm{N}_{0}$ After 4 minutes, $2500=\lambda \mathrm{N}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-4 \lambda}$ $\mathrm{N} \lambda=\mathrm{N}_{0} \lambda \mathrm{e}^{-4 \lambda}$ $2500=10000 \mathrm{e}^{-4 \lambda}$ $\mathrm{e}^{-4 \lambda}=\frac{1}{4}$ Taking natural $\log$ both side, we get- $\ln \mathrm{e}^{-4 \lambda}=\ln \left(\frac{1}{4}\right)$ $-4 \lambda \ln =\ln \left(2^{-2}\right)$ $-4 \lambda=-2 \ln 2$ $\lambda=0.5 \ln 2$ $\lambda=0.5 \log _{\mathrm{e}} 2$
MHT-CET 2017
NUCLEAR PHYSICS
147836
The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t=0$ and $\mathrm{N}_{0} / \mathrm{e}$ counts per minute at $\mathrm{t}=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
1 $\log _{\mathrm{e}} 2 / 5$
2 $\frac{5}{\log _{\mathrm{e}} 2}$
3 $5 \log 10^{2}$
4 $5 \log _{\mathrm{e}} 2$
Explanation:
D Given that, $\mathrm{R}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}}$ count per minute $\mathrm{R}=\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}$ count per minute $\mathrm{t}=5$ minute From activity law, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-5 \lambda}$ $\mathrm{e}^{-1}=\mathrm{e}^{-5 \lambda}$ $5 \lambda=1$ $\lambda=1 / 5 \text { per minute }$ At, $t=T_{1 / 2}=$ The activity $R$ reduces to $R_{0} / 2$ From equation (i), we get - $\frac{\mathrm{R}_{\mathrm{o}}}{2}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $2=\mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ Logarithms of both sides, we get - $\lambda \mathrm{T}_{1 / 2}=\log _{\mathrm{e}} 2$ $\mathrm{~T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}=\frac{\log _{\mathrm{e}} 2}{1 / 5}$ $\mathrm{~T}_{1 / 2}=5 \log _{\mathrm{e}} 2 \text { minutes } \quad\left[\because \lambda=\frac{1}{5}\right]$
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 $t$ $-$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Where, $\mathrm{N}_{0}=$ initial number of nuclide This equation is equivalent, $y=a e^{-k x}$ Hence, the radioactive element A decay in stable element $\mathrm{B}$ 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 $\left(\mathrm{T}_{1 / 2}\right)=2.3$ day $\lambda=$ ? We know that, Decay constant $(\lambda)=\left(\frac{0.693}{T_{1 / 2}}\right)$ $\lambda=\frac{0.693}{2.3}$ $\lambda=\frac{6930}{23000}$ $\lambda=0.3$
CG PET- 2007
NUCLEAR PHYSICS
147839
If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2}=5 \mathrm{yrs}$, then the amount of substance left after $15 \mathrm{yr}$ is
1 $\frac{\mathrm{N}_{0}}{8}$
2 $\frac{\mathrm{N}_{0}}{16}$
3 $\frac{\mathrm{N}_{0}}{2}$
4 $\frac{\mathrm{N}_{0}}{4}$
Explanation:
A Given that, Original mass of substance $=\mathrm{N}_{\mathrm{o}}$ $\mathrm{T}_{1 / 2}=5$ year Time $(\mathrm{t})=15 \mathrm{yrs}$ Amount left after decay is given by $\mathrm{N}$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Number of half-lives $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ Putting the value of $n$, we get - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$
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 $2 \log _{\mathrm{e}} 2$
2 $0.5 \log _{\mathrm{e}} 2$
3 $0.6 \log _{\mathrm{e}} 2$
4 $0.8 \log _{\mathrm{e}} 2$
Explanation:
B Given that, $\frac{\mathrm{dN}}{\mathrm{dt}}=10000 \text { per minuts }$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}_{0}$ $10000=\lambda \mathrm{N}_{0}$ After 4 minutes, $2500=\lambda \mathrm{N}$ We know that, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-4 \lambda}$ $\mathrm{N} \lambda=\mathrm{N}_{0} \lambda \mathrm{e}^{-4 \lambda}$ $2500=10000 \mathrm{e}^{-4 \lambda}$ $\mathrm{e}^{-4 \lambda}=\frac{1}{4}$ Taking natural $\log$ both side, we get- $\ln \mathrm{e}^{-4 \lambda}=\ln \left(\frac{1}{4}\right)$ $-4 \lambda \ln =\ln \left(2^{-2}\right)$ $-4 \lambda=-2 \ln 2$ $\lambda=0.5 \ln 2$ $\lambda=0.5 \log _{\mathrm{e}} 2$
MHT-CET 2017
NUCLEAR PHYSICS
147836
The activity of a radioactive sample is measured as $N_{0}$ counts per minute at $t=0$ and $\mathrm{N}_{0} / \mathrm{e}$ counts per minute at $\mathrm{t}=5$ minutes. The time (in minutes) at which the activity reduces to half its value is
1 $\log _{\mathrm{e}} 2 / 5$
2 $\frac{5}{\log _{\mathrm{e}} 2}$
3 $5 \log 10^{2}$
4 $5 \log _{\mathrm{e}} 2$
Explanation:
D Given that, $\mathrm{R}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}}$ count per minute $\mathrm{R}=\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}$ count per minute $\mathrm{t}=5$ minute From activity law, $\mathrm{R}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda t}$ $\frac{\mathrm{N}_{\mathrm{o}}}{\mathrm{e}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-5 \lambda}$ $\mathrm{e}^{-1}=\mathrm{e}^{-5 \lambda}$ $5 \lambda=1$ $\lambda=1 / 5 \text { per minute }$ At, $t=T_{1 / 2}=$ The activity $R$ reduces to $R_{0} / 2$ From equation (i), we get - $\frac{\mathrm{R}_{\mathrm{o}}}{2}=\mathrm{R}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $2=\mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ Logarithms of both sides, we get - $\lambda \mathrm{T}_{1 / 2}=\log _{\mathrm{e}} 2$ $\mathrm{~T}_{1 / 2}=\frac{\log _{\mathrm{e}} 2}{\lambda}=\frac{\log _{\mathrm{e}} 2}{1 / 5}$ $\mathrm{~T}_{1 / 2}=5 \log _{\mathrm{e}} 2 \text { minutes } \quad\left[\because \lambda=\frac{1}{5}\right]$
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 $t$ $-$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Where, $\mathrm{N}_{0}=$ initial number of nuclide This equation is equivalent, $y=a e^{-k x}$ Hence, the radioactive element A decay in stable element $\mathrm{B}$ 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 $\left(\mathrm{T}_{1 / 2}\right)=2.3$ day $\lambda=$ ? We know that, Decay constant $(\lambda)=\left(\frac{0.693}{T_{1 / 2}}\right)$ $\lambda=\frac{0.693}{2.3}$ $\lambda=\frac{6930}{23000}$ $\lambda=0.3$
CG PET- 2007
NUCLEAR PHYSICS
147839
If $N_{0}$ is the original mass of the substance of half-life period $T_{1 / 2}=5 \mathrm{yrs}$, then the amount of substance left after $15 \mathrm{yr}$ is
1 $\frac{\mathrm{N}_{0}}{8}$
2 $\frac{\mathrm{N}_{0}}{16}$
3 $\frac{\mathrm{N}_{0}}{2}$
4 $\frac{\mathrm{N}_{0}}{4}$
Explanation:
A Given that, Original mass of substance $=\mathrm{N}_{\mathrm{o}}$ $\mathrm{T}_{1 / 2}=5$ year Time $(\mathrm{t})=15 \mathrm{yrs}$ Amount left after decay is given by $\mathrm{N}$, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Number of half-lives $(\mathrm{n})=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{15}{5}=3$ Putting the value of $n$, we get - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{3}$ $\mathrm{~N}=\frac{\mathrm{N}_{0}}{8}$
