147726
Which one of the following graphs represents the graph between the instantaneous the graph between the instantaneous concentration $N$ of a radioactive element and time $t$ ?
1 a
2 b
3 c
4 d
Explanation:
B According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ It shows that $\mathrm{N}$ decrease exponentially with time.
BCECE-2012
NUCLEAR PHYSICS
147727
A radioactive substance has half-life of $60 \mathrm{~min}$ During $3 \mathrm{~h}$, the fraction of the substance that has to be decayed, will be-
1 $87.5 \%$
2 $52.5 \%$
3 $25.5 \%$
4 $8.5 \%$
Explanation:
A The rate of decay of a radioactive substance is proportional to number of atoms left at that instant $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Where, $\mathrm{N}_{\mathrm{o}}=$ original number of atoms $\mathrm{n}=$ number of half lives $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{180}{60}=3$ $\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$ $\therefore \quad \mathrm{N}=\frac{\mathrm{N}_{\mathrm{o}}}{8}=0.125 \mathrm{~N}_{\mathrm{o}}=12.5 \% \mathrm{~N}$ Amount decayed $=100-12.5=87.5 \%$
BCECE-2007
NUCLEAR PHYSICS
147729
If half-life of radium is 77 days, its decay constant will be :
1 $3 \times 10^{-3} /$ day
2 $9 \times 10^{-3} /$ day
3 $1 \times 10^{-3} /$ day
4 $6 \times 10^{-3} /$ day
Explanation:
B The time required for the number of parent nuclei to fall $50 \%$ is called half-life $\mathrm{T}_{1 / 2}$ and may be related to $\lambda$ as follows $0.5 \mathrm{~N}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $\lambda \mathrm{T}_{1 / 2}=\ln (2)=0.693$ $\mathrm{T}_{1 / 2}=\frac{0.693}{\lambda}$ $\lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}$ $\because \mathrm{T}_{1 / 2}=77 \text { days }$ $\lambda=\frac{0.693}{77}=9 \times 10^{-3} / \text { days }$
BCECE-2005
NUCLEAR PHYSICS
147731
Nuclear reactor in which $\mathrm{U}-235$ is used as fuel. Uses $2 \mathrm{~kg}$ of U-235 in 30 days. Then, power output of the reactor will be (given energy released per fission $=185 \mathrm{Me} \mathrm{V}$ )
1 $43.5 \mathrm{MW}$
2 $58.5 \mathrm{MW}$
3 $69.6 \mathrm{MW}$
4 $73.1 \mathrm{MW}$
Explanation:
B Number of atoms in $2 \mathrm{~kg}$ of uranium $=\frac{6.02 \times 10^{23}}{235} \times 2000$ $=5.12 \times 10^{24}$ Energy obtained from these atoms is $=5.12 \times 10^{24} \times 185 \mathrm{meV}$ $=5.12 \times 10^{24} \times 185 \times 10^{6} \mathrm{eV}$ Energy obtained per second is $=\frac{5.12 \times 10^{24} \times 185 \times 10^{6} \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60}$ $=58.47 \times 10^{6} \mathrm{~W}$ $=58.5 \mathrm{MW}$
147726
Which one of the following graphs represents the graph between the instantaneous the graph between the instantaneous concentration $N$ of a radioactive element and time $t$ ?
1 a
2 b
3 c
4 d
Explanation:
B According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ It shows that $\mathrm{N}$ decrease exponentially with time.
BCECE-2012
NUCLEAR PHYSICS
147727
A radioactive substance has half-life of $60 \mathrm{~min}$ During $3 \mathrm{~h}$, the fraction of the substance that has to be decayed, will be-
1 $87.5 \%$
2 $52.5 \%$
3 $25.5 \%$
4 $8.5 \%$
Explanation:
A The rate of decay of a radioactive substance is proportional to number of atoms left at that instant $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Where, $\mathrm{N}_{\mathrm{o}}=$ original number of atoms $\mathrm{n}=$ number of half lives $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{180}{60}=3$ $\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$ $\therefore \quad \mathrm{N}=\frac{\mathrm{N}_{\mathrm{o}}}{8}=0.125 \mathrm{~N}_{\mathrm{o}}=12.5 \% \mathrm{~N}$ Amount decayed $=100-12.5=87.5 \%$
BCECE-2007
NUCLEAR PHYSICS
147729
If half-life of radium is 77 days, its decay constant will be :
1 $3 \times 10^{-3} /$ day
2 $9 \times 10^{-3} /$ day
3 $1 \times 10^{-3} /$ day
4 $6 \times 10^{-3} /$ day
Explanation:
B The time required for the number of parent nuclei to fall $50 \%$ is called half-life $\mathrm{T}_{1 / 2}$ and may be related to $\lambda$ as follows $0.5 \mathrm{~N}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $\lambda \mathrm{T}_{1 / 2}=\ln (2)=0.693$ $\mathrm{T}_{1 / 2}=\frac{0.693}{\lambda}$ $\lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}$ $\because \mathrm{T}_{1 / 2}=77 \text { days }$ $\lambda=\frac{0.693}{77}=9 \times 10^{-3} / \text { days }$
BCECE-2005
NUCLEAR PHYSICS
147731
Nuclear reactor in which $\mathrm{U}-235$ is used as fuel. Uses $2 \mathrm{~kg}$ of U-235 in 30 days. Then, power output of the reactor will be (given energy released per fission $=185 \mathrm{Me} \mathrm{V}$ )
1 $43.5 \mathrm{MW}$
2 $58.5 \mathrm{MW}$
3 $69.6 \mathrm{MW}$
4 $73.1 \mathrm{MW}$
Explanation:
B Number of atoms in $2 \mathrm{~kg}$ of uranium $=\frac{6.02 \times 10^{23}}{235} \times 2000$ $=5.12 \times 10^{24}$ Energy obtained from these atoms is $=5.12 \times 10^{24} \times 185 \mathrm{meV}$ $=5.12 \times 10^{24} \times 185 \times 10^{6} \mathrm{eV}$ Energy obtained per second is $=\frac{5.12 \times 10^{24} \times 185 \times 10^{6} \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60}$ $=58.47 \times 10^{6} \mathrm{~W}$ $=58.5 \mathrm{MW}$
147726
Which one of the following graphs represents the graph between the instantaneous the graph between the instantaneous concentration $N$ of a radioactive element and time $t$ ?
1 a
2 b
3 c
4 d
Explanation:
B According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ It shows that $\mathrm{N}$ decrease exponentially with time.
BCECE-2012
NUCLEAR PHYSICS
147727
A radioactive substance has half-life of $60 \mathrm{~min}$ During $3 \mathrm{~h}$, the fraction of the substance that has to be decayed, will be-
1 $87.5 \%$
2 $52.5 \%$
3 $25.5 \%$
4 $8.5 \%$
Explanation:
A The rate of decay of a radioactive substance is proportional to number of atoms left at that instant $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Where, $\mathrm{N}_{\mathrm{o}}=$ original number of atoms $\mathrm{n}=$ number of half lives $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{180}{60}=3$ $\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$ $\therefore \quad \mathrm{N}=\frac{\mathrm{N}_{\mathrm{o}}}{8}=0.125 \mathrm{~N}_{\mathrm{o}}=12.5 \% \mathrm{~N}$ Amount decayed $=100-12.5=87.5 \%$
BCECE-2007
NUCLEAR PHYSICS
147729
If half-life of radium is 77 days, its decay constant will be :
1 $3 \times 10^{-3} /$ day
2 $9 \times 10^{-3} /$ day
3 $1 \times 10^{-3} /$ day
4 $6 \times 10^{-3} /$ day
Explanation:
B The time required for the number of parent nuclei to fall $50 \%$ is called half-life $\mathrm{T}_{1 / 2}$ and may be related to $\lambda$ as follows $0.5 \mathrm{~N}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $\lambda \mathrm{T}_{1 / 2}=\ln (2)=0.693$ $\mathrm{T}_{1 / 2}=\frac{0.693}{\lambda}$ $\lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}$ $\because \mathrm{T}_{1 / 2}=77 \text { days }$ $\lambda=\frac{0.693}{77}=9 \times 10^{-3} / \text { days }$
BCECE-2005
NUCLEAR PHYSICS
147731
Nuclear reactor in which $\mathrm{U}-235$ is used as fuel. Uses $2 \mathrm{~kg}$ of U-235 in 30 days. Then, power output of the reactor will be (given energy released per fission $=185 \mathrm{Me} \mathrm{V}$ )
1 $43.5 \mathrm{MW}$
2 $58.5 \mathrm{MW}$
3 $69.6 \mathrm{MW}$
4 $73.1 \mathrm{MW}$
Explanation:
B Number of atoms in $2 \mathrm{~kg}$ of uranium $=\frac{6.02 \times 10^{23}}{235} \times 2000$ $=5.12 \times 10^{24}$ Energy obtained from these atoms is $=5.12 \times 10^{24} \times 185 \mathrm{meV}$ $=5.12 \times 10^{24} \times 185 \times 10^{6} \mathrm{eV}$ Energy obtained per second is $=\frac{5.12 \times 10^{24} \times 185 \times 10^{6} \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60}$ $=58.47 \times 10^{6} \mathrm{~W}$ $=58.5 \mathrm{MW}$
147726
Which one of the following graphs represents the graph between the instantaneous the graph between the instantaneous concentration $N$ of a radioactive element and time $t$ ?
1 a
2 b
3 c
4 d
Explanation:
B According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ It shows that $\mathrm{N}$ decrease exponentially with time.
BCECE-2012
NUCLEAR PHYSICS
147727
A radioactive substance has half-life of $60 \mathrm{~min}$ During $3 \mathrm{~h}$, the fraction of the substance that has to be decayed, will be-
1 $87.5 \%$
2 $52.5 \%$
3 $25.5 \%$
4 $8.5 \%$
Explanation:
A The rate of decay of a radioactive substance is proportional to number of atoms left at that instant $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\frac{1}{2}\right)^{\mathrm{n}}$ Where, $\mathrm{N}_{\mathrm{o}}=$ original number of atoms $\mathrm{n}=$ number of half lives $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{180}{60}=3$ $\therefore \quad \frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$ $\therefore \quad \mathrm{N}=\frac{\mathrm{N}_{\mathrm{o}}}{8}=0.125 \mathrm{~N}_{\mathrm{o}}=12.5 \% \mathrm{~N}$ Amount decayed $=100-12.5=87.5 \%$
BCECE-2007
NUCLEAR PHYSICS
147729
If half-life of radium is 77 days, its decay constant will be :
1 $3 \times 10^{-3} /$ day
2 $9 \times 10^{-3} /$ day
3 $1 \times 10^{-3} /$ day
4 $6 \times 10^{-3} /$ day
Explanation:
B The time required for the number of parent nuclei to fall $50 \%$ is called half-life $\mathrm{T}_{1 / 2}$ and may be related to $\lambda$ as follows $0.5 \mathrm{~N}_{\mathrm{o}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{T}_{1 / 2}}$ $\lambda \mathrm{T}_{1 / 2}=\ln (2)=0.693$ $\mathrm{T}_{1 / 2}=\frac{0.693}{\lambda}$ $\lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}$ $\because \mathrm{T}_{1 / 2}=77 \text { days }$ $\lambda=\frac{0.693}{77}=9 \times 10^{-3} / \text { days }$
BCECE-2005
NUCLEAR PHYSICS
147731
Nuclear reactor in which $\mathrm{U}-235$ is used as fuel. Uses $2 \mathrm{~kg}$ of U-235 in 30 days. Then, power output of the reactor will be (given energy released per fission $=185 \mathrm{Me} \mathrm{V}$ )
1 $43.5 \mathrm{MW}$
2 $58.5 \mathrm{MW}$
3 $69.6 \mathrm{MW}$
4 $73.1 \mathrm{MW}$
Explanation:
B Number of atoms in $2 \mathrm{~kg}$ of uranium $=\frac{6.02 \times 10^{23}}{235} \times 2000$ $=5.12 \times 10^{24}$ Energy obtained from these atoms is $=5.12 \times 10^{24} \times 185 \mathrm{meV}$ $=5.12 \times 10^{24} \times 185 \times 10^{6} \mathrm{eV}$ Energy obtained per second is $=\frac{5.12 \times 10^{24} \times 185 \times 10^{6} \times 1.6 \times 10^{-19}}{30 \times 24 \times 60 \times 60}$ $=58.47 \times 10^{6} \mathrm{~W}$ $=58.5 \mathrm{MW}$
