147605
For an radioactive element, $\tau=$ $\tau_{\underline{1}}$
1 0.693
2 693
3 144
4 1.44
Explanation:
D Let, Half life of radioactive element $=\tau_{1 / 2}$ Mean life-time $=\tau$ So, $\tau=\frac{1}{\lambda}$ $\tau=\frac{1}{\ln 2} \times \tau_{1 / 2} \quad\left(\because \lambda=\frac{\ln 2}{\tau_{1 / 2}}\right)$ $\tau=\frac{\tau_{1 / 2}}{0.693}$ $\tau=1.44 \tau_{1 / 2}$
GUJCET 2019
NUCLEAR PHYSICS
147607
A radioactive element $X$ converts into another stable element $Y$. half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is
1 2 hours
2 4 hours
3 Between 4 hours and 6 hours
4 6 hours
Explanation:
C Assume initial composition to contain only. Given, half life of $X=2$ hour After one half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{2} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{1}{2} \mathrm{~N}$ Hence, ratio is $1: 1$ After two half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{4} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{3}{4} \mathrm{~N}$ Ratio is $1: 3$ After three half lives - Amount of $\mathrm{X}$ remaining is $=\frac{1}{8} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{7}{8} \mathrm{~N}$ Hence ratio is $1: 7$ The ratio of $1: 4$ is obtained somewhere between two and three half lives. Since each half life is 2 hours long somewhere between 4 to 6 hours the ratio of $1: 4$ is obtained.
AP EAMCET (20.04.2019) Shift-1
NUCLEAR PHYSICS
147608
The rate of disintegration of a radioactive sample is $R$ and the number of atoms present at any time $t$ is $N$. When $\frac{R}{N}$ is taken along $Y$ axis and $t$ is taken along $\mathrm{X}$-axis, the correct graphs is
1 a
2 b
3 c
4 d
Explanation:
D Law of radioactive decay is given as, $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Where, $\frac{\mathrm{dN}}{\mathrm{dt}}=$ Rate of disintegration So, $\mathrm{R}=-\lambda \mathrm{N} \Rightarrow \frac{\mathrm{R}}{\mathrm{N}}=-\lambda=\mathrm{constant}$ Where, $\lambda=$ decay constant So, the graph $\frac{\mathrm{R}}{\mathrm{N}}$ versus $\mathrm{t}$ is as straight line parallel to $\mathrm{X}$ axis. So, the correct graph shown in option (d).
AP EAMCET (21.04.2019) Shift-I
NUCLEAR PHYSICS
147609
Half - life of a radioactive substance is $\mathbf{1 8}$ minutes. The time interval between its $20 \%$ decay and $\mathbf{8 0} \%$ decay in minutes is
1 6
2 9
3 18
4 36
Explanation:
D Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=18 \mathrm{~min}$ After $n$ half-life the numbers of atom left undecayed is given by- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ For $20 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{20}{100} \mathrm{~N}_{0}=0.8 \mathrm{~N}_{0}$ $0.8 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}}{18}}$ For $80 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{80}{100} \mathrm{~N}_{0}=0.2 \mathrm{~N}_{0}$ $0.2 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{2}}{18}}$ On dividing equation (i) and (ii), we get- $4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$ Taking $\log$ on the both sides, we get- $\log (4)=\log \left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{18}}$ $\log (2)^{2}=\log (2)^{\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}}$ $2=\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}$ $36=\mathrm{t}_{2}-\mathrm{t}_{1}$ So, $\quad \mathrm{t}_{2}-\mathrm{t}_{1}=36$
AP EAMCET (20.04.2019) Shift-II
NUCLEAR PHYSICS
147610
In a nuclear reactor the activity of a radioactive substance is $2000/ \mathrm{s}$. If the mean life of the products is $\mathbf{5 0}$ minutes, then in the steady power generation, the number of radio nuclides is
1 $12 \times 10^{5}$
2 $60 \times 10^{5}$
3 $90 \times 10^{5}$
4 $15 \times 10^{5}$
Explanation:
B Given, $\frac{\mathrm{dN}}{\mathrm{dt}}=2000 / \mathrm{s}$ Mean life $(\tau)=50 \mathrm{~min}=50 \times 60 \mathrm{sec}$ Since, we know that mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$. $\therefore \quad \tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }$ The rate of decay is proportional to the number of radio nuclides is given as, $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $2000=\frac{1}{50 \times 60} \times \mathrm{N}$ $\mathrm{N}=2000\times 50 \times 60$ $\mathrm{~N}=60 \times 10^{5}$ Hence, the number of radio nuclides is $60 \times 10^{5}$.
147605
For an radioactive element, $\tau=$ $\tau_{\underline{1}}$
1 0.693
2 693
3 144
4 1.44
Explanation:
D Let, Half life of radioactive element $=\tau_{1 / 2}$ Mean life-time $=\tau$ So, $\tau=\frac{1}{\lambda}$ $\tau=\frac{1}{\ln 2} \times \tau_{1 / 2} \quad\left(\because \lambda=\frac{\ln 2}{\tau_{1 / 2}}\right)$ $\tau=\frac{\tau_{1 / 2}}{0.693}$ $\tau=1.44 \tau_{1 / 2}$
GUJCET 2019
NUCLEAR PHYSICS
147607
A radioactive element $X$ converts into another stable element $Y$. half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is
1 2 hours
2 4 hours
3 Between 4 hours and 6 hours
4 6 hours
Explanation:
C Assume initial composition to contain only. Given, half life of $X=2$ hour After one half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{2} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{1}{2} \mathrm{~N}$ Hence, ratio is $1: 1$ After two half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{4} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{3}{4} \mathrm{~N}$ Ratio is $1: 3$ After three half lives - Amount of $\mathrm{X}$ remaining is $=\frac{1}{8} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{7}{8} \mathrm{~N}$ Hence ratio is $1: 7$ The ratio of $1: 4$ is obtained somewhere between two and three half lives. Since each half life is 2 hours long somewhere between 4 to 6 hours the ratio of $1: 4$ is obtained.
AP EAMCET (20.04.2019) Shift-1
NUCLEAR PHYSICS
147608
The rate of disintegration of a radioactive sample is $R$ and the number of atoms present at any time $t$ is $N$. When $\frac{R}{N}$ is taken along $Y$ axis and $t$ is taken along $\mathrm{X}$-axis, the correct graphs is
1 a
2 b
3 c
4 d
Explanation:
D Law of radioactive decay is given as, $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Where, $\frac{\mathrm{dN}}{\mathrm{dt}}=$ Rate of disintegration So, $\mathrm{R}=-\lambda \mathrm{N} \Rightarrow \frac{\mathrm{R}}{\mathrm{N}}=-\lambda=\mathrm{constant}$ Where, $\lambda=$ decay constant So, the graph $\frac{\mathrm{R}}{\mathrm{N}}$ versus $\mathrm{t}$ is as straight line parallel to $\mathrm{X}$ axis. So, the correct graph shown in option (d).
AP EAMCET (21.04.2019) Shift-I
NUCLEAR PHYSICS
147609
Half - life of a radioactive substance is $\mathbf{1 8}$ minutes. The time interval between its $20 \%$ decay and $\mathbf{8 0} \%$ decay in minutes is
1 6
2 9
3 18
4 36
Explanation:
D Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=18 \mathrm{~min}$ After $n$ half-life the numbers of atom left undecayed is given by- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ For $20 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{20}{100} \mathrm{~N}_{0}=0.8 \mathrm{~N}_{0}$ $0.8 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}}{18}}$ For $80 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{80}{100} \mathrm{~N}_{0}=0.2 \mathrm{~N}_{0}$ $0.2 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{2}}{18}}$ On dividing equation (i) and (ii), we get- $4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$ Taking $\log$ on the both sides, we get- $\log (4)=\log \left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{18}}$ $\log (2)^{2}=\log (2)^{\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}}$ $2=\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}$ $36=\mathrm{t}_{2}-\mathrm{t}_{1}$ So, $\quad \mathrm{t}_{2}-\mathrm{t}_{1}=36$
AP EAMCET (20.04.2019) Shift-II
NUCLEAR PHYSICS
147610
In a nuclear reactor the activity of a radioactive substance is $2000/ \mathrm{s}$. If the mean life of the products is $\mathbf{5 0}$ minutes, then in the steady power generation, the number of radio nuclides is
1 $12 \times 10^{5}$
2 $60 \times 10^{5}$
3 $90 \times 10^{5}$
4 $15 \times 10^{5}$
Explanation:
B Given, $\frac{\mathrm{dN}}{\mathrm{dt}}=2000 / \mathrm{s}$ Mean life $(\tau)=50 \mathrm{~min}=50 \times 60 \mathrm{sec}$ Since, we know that mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$. $\therefore \quad \tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }$ The rate of decay is proportional to the number of radio nuclides is given as, $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $2000=\frac{1}{50 \times 60} \times \mathrm{N}$ $\mathrm{N}=2000\times 50 \times 60$ $\mathrm{~N}=60 \times 10^{5}$ Hence, the number of radio nuclides is $60 \times 10^{5}$.
147605
For an radioactive element, $\tau=$ $\tau_{\underline{1}}$
1 0.693
2 693
3 144
4 1.44
Explanation:
D Let, Half life of radioactive element $=\tau_{1 / 2}$ Mean life-time $=\tau$ So, $\tau=\frac{1}{\lambda}$ $\tau=\frac{1}{\ln 2} \times \tau_{1 / 2} \quad\left(\because \lambda=\frac{\ln 2}{\tau_{1 / 2}}\right)$ $\tau=\frac{\tau_{1 / 2}}{0.693}$ $\tau=1.44 \tau_{1 / 2}$
GUJCET 2019
NUCLEAR PHYSICS
147607
A radioactive element $X$ converts into another stable element $Y$. half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is
1 2 hours
2 4 hours
3 Between 4 hours and 6 hours
4 6 hours
Explanation:
C Assume initial composition to contain only. Given, half life of $X=2$ hour After one half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{2} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{1}{2} \mathrm{~N}$ Hence, ratio is $1: 1$ After two half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{4} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{3}{4} \mathrm{~N}$ Ratio is $1: 3$ After three half lives - Amount of $\mathrm{X}$ remaining is $=\frac{1}{8} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{7}{8} \mathrm{~N}$ Hence ratio is $1: 7$ The ratio of $1: 4$ is obtained somewhere between two and three half lives. Since each half life is 2 hours long somewhere between 4 to 6 hours the ratio of $1: 4$ is obtained.
AP EAMCET (20.04.2019) Shift-1
NUCLEAR PHYSICS
147608
The rate of disintegration of a radioactive sample is $R$ and the number of atoms present at any time $t$ is $N$. When $\frac{R}{N}$ is taken along $Y$ axis and $t$ is taken along $\mathrm{X}$-axis, the correct graphs is
1 a
2 b
3 c
4 d
Explanation:
D Law of radioactive decay is given as, $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Where, $\frac{\mathrm{dN}}{\mathrm{dt}}=$ Rate of disintegration So, $\mathrm{R}=-\lambda \mathrm{N} \Rightarrow \frac{\mathrm{R}}{\mathrm{N}}=-\lambda=\mathrm{constant}$ Where, $\lambda=$ decay constant So, the graph $\frac{\mathrm{R}}{\mathrm{N}}$ versus $\mathrm{t}$ is as straight line parallel to $\mathrm{X}$ axis. So, the correct graph shown in option (d).
AP EAMCET (21.04.2019) Shift-I
NUCLEAR PHYSICS
147609
Half - life of a radioactive substance is $\mathbf{1 8}$ minutes. The time interval between its $20 \%$ decay and $\mathbf{8 0} \%$ decay in minutes is
1 6
2 9
3 18
4 36
Explanation:
D Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=18 \mathrm{~min}$ After $n$ half-life the numbers of atom left undecayed is given by- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ For $20 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{20}{100} \mathrm{~N}_{0}=0.8 \mathrm{~N}_{0}$ $0.8 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}}{18}}$ For $80 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{80}{100} \mathrm{~N}_{0}=0.2 \mathrm{~N}_{0}$ $0.2 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{2}}{18}}$ On dividing equation (i) and (ii), we get- $4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$ Taking $\log$ on the both sides, we get- $\log (4)=\log \left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{18}}$ $\log (2)^{2}=\log (2)^{\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}}$ $2=\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}$ $36=\mathrm{t}_{2}-\mathrm{t}_{1}$ So, $\quad \mathrm{t}_{2}-\mathrm{t}_{1}=36$
AP EAMCET (20.04.2019) Shift-II
NUCLEAR PHYSICS
147610
In a nuclear reactor the activity of a radioactive substance is $2000/ \mathrm{s}$. If the mean life of the products is $\mathbf{5 0}$ minutes, then in the steady power generation, the number of radio nuclides is
1 $12 \times 10^{5}$
2 $60 \times 10^{5}$
3 $90 \times 10^{5}$
4 $15 \times 10^{5}$
Explanation:
B Given, $\frac{\mathrm{dN}}{\mathrm{dt}}=2000 / \mathrm{s}$ Mean life $(\tau)=50 \mathrm{~min}=50 \times 60 \mathrm{sec}$ Since, we know that mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$. $\therefore \quad \tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }$ The rate of decay is proportional to the number of radio nuclides is given as, $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $2000=\frac{1}{50 \times 60} \times \mathrm{N}$ $\mathrm{N}=2000\times 50 \times 60$ $\mathrm{~N}=60 \times 10^{5}$ Hence, the number of radio nuclides is $60 \times 10^{5}$.
147605
For an radioactive element, $\tau=$ $\tau_{\underline{1}}$
1 0.693
2 693
3 144
4 1.44
Explanation:
D Let, Half life of radioactive element $=\tau_{1 / 2}$ Mean life-time $=\tau$ So, $\tau=\frac{1}{\lambda}$ $\tau=\frac{1}{\ln 2} \times \tau_{1 / 2} \quad\left(\because \lambda=\frac{\ln 2}{\tau_{1 / 2}}\right)$ $\tau=\frac{\tau_{1 / 2}}{0.693}$ $\tau=1.44 \tau_{1 / 2}$
GUJCET 2019
NUCLEAR PHYSICS
147607
A radioactive element $X$ converts into another stable element $Y$. half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is
1 2 hours
2 4 hours
3 Between 4 hours and 6 hours
4 6 hours
Explanation:
C Assume initial composition to contain only. Given, half life of $X=2$ hour After one half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{2} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{1}{2} \mathrm{~N}$ Hence, ratio is $1: 1$ After two half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{4} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{3}{4} \mathrm{~N}$ Ratio is $1: 3$ After three half lives - Amount of $\mathrm{X}$ remaining is $=\frac{1}{8} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{7}{8} \mathrm{~N}$ Hence ratio is $1: 7$ The ratio of $1: 4$ is obtained somewhere between two and three half lives. Since each half life is 2 hours long somewhere between 4 to 6 hours the ratio of $1: 4$ is obtained.
AP EAMCET (20.04.2019) Shift-1
NUCLEAR PHYSICS
147608
The rate of disintegration of a radioactive sample is $R$ and the number of atoms present at any time $t$ is $N$. When $\frac{R}{N}$ is taken along $Y$ axis and $t$ is taken along $\mathrm{X}$-axis, the correct graphs is
1 a
2 b
3 c
4 d
Explanation:
D Law of radioactive decay is given as, $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Where, $\frac{\mathrm{dN}}{\mathrm{dt}}=$ Rate of disintegration So, $\mathrm{R}=-\lambda \mathrm{N} \Rightarrow \frac{\mathrm{R}}{\mathrm{N}}=-\lambda=\mathrm{constant}$ Where, $\lambda=$ decay constant So, the graph $\frac{\mathrm{R}}{\mathrm{N}}$ versus $\mathrm{t}$ is as straight line parallel to $\mathrm{X}$ axis. So, the correct graph shown in option (d).
AP EAMCET (21.04.2019) Shift-I
NUCLEAR PHYSICS
147609
Half - life of a radioactive substance is $\mathbf{1 8}$ minutes. The time interval between its $20 \%$ decay and $\mathbf{8 0} \%$ decay in minutes is
1 6
2 9
3 18
4 36
Explanation:
D Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=18 \mathrm{~min}$ After $n$ half-life the numbers of atom left undecayed is given by- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ For $20 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{20}{100} \mathrm{~N}_{0}=0.8 \mathrm{~N}_{0}$ $0.8 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}}{18}}$ For $80 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{80}{100} \mathrm{~N}_{0}=0.2 \mathrm{~N}_{0}$ $0.2 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{2}}{18}}$ On dividing equation (i) and (ii), we get- $4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$ Taking $\log$ on the both sides, we get- $\log (4)=\log \left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{18}}$ $\log (2)^{2}=\log (2)^{\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}}$ $2=\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}$ $36=\mathrm{t}_{2}-\mathrm{t}_{1}$ So, $\quad \mathrm{t}_{2}-\mathrm{t}_{1}=36$
AP EAMCET (20.04.2019) Shift-II
NUCLEAR PHYSICS
147610
In a nuclear reactor the activity of a radioactive substance is $2000/ \mathrm{s}$. If the mean life of the products is $\mathbf{5 0}$ minutes, then in the steady power generation, the number of radio nuclides is
1 $12 \times 10^{5}$
2 $60 \times 10^{5}$
3 $90 \times 10^{5}$
4 $15 \times 10^{5}$
Explanation:
B Given, $\frac{\mathrm{dN}}{\mathrm{dt}}=2000 / \mathrm{s}$ Mean life $(\tau)=50 \mathrm{~min}=50 \times 60 \mathrm{sec}$ Since, we know that mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$. $\therefore \quad \tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }$ The rate of decay is proportional to the number of radio nuclides is given as, $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $2000=\frac{1}{50 \times 60} \times \mathrm{N}$ $\mathrm{N}=2000\times 50 \times 60$ $\mathrm{~N}=60 \times 10^{5}$ Hence, the number of radio nuclides is $60 \times 10^{5}$.
147605
For an radioactive element, $\tau=$ $\tau_{\underline{1}}$
1 0.693
2 693
3 144
4 1.44
Explanation:
D Let, Half life of radioactive element $=\tau_{1 / 2}$ Mean life-time $=\tau$ So, $\tau=\frac{1}{\lambda}$ $\tau=\frac{1}{\ln 2} \times \tau_{1 / 2} \quad\left(\because \lambda=\frac{\ln 2}{\tau_{1 / 2}}\right)$ $\tau=\frac{\tau_{1 / 2}}{0.693}$ $\tau=1.44 \tau_{1 / 2}$
GUJCET 2019
NUCLEAR PHYSICS
147607
A radioactive element $X$ converts into another stable element $Y$. half life of $X$ is 2 hours. Initially only $X$ is present. After a time $t$, if the ratio of atoms of $X$ to $Y$ is $1: 4$, then the value of $t$ is
1 2 hours
2 4 hours
3 Between 4 hours and 6 hours
4 6 hours
Explanation:
C Assume initial composition to contain only. Given, half life of $X=2$ hour After one half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{2} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{1}{2} \mathrm{~N}$ Hence, ratio is $1: 1$ After two half life - Amount of $\mathrm{X}$ remaining is $=\frac{1}{4} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{3}{4} \mathrm{~N}$ Ratio is $1: 3$ After three half lives - Amount of $\mathrm{X}$ remaining is $=\frac{1}{8} \mathrm{~N}$ Amount of $\mathrm{Y}$ is $=\frac{7}{8} \mathrm{~N}$ Hence ratio is $1: 7$ The ratio of $1: 4$ is obtained somewhere between two and three half lives. Since each half life is 2 hours long somewhere between 4 to 6 hours the ratio of $1: 4$ is obtained.
AP EAMCET (20.04.2019) Shift-1
NUCLEAR PHYSICS
147608
The rate of disintegration of a radioactive sample is $R$ and the number of atoms present at any time $t$ is $N$. When $\frac{R}{N}$ is taken along $Y$ axis and $t$ is taken along $\mathrm{X}$-axis, the correct graphs is
1 a
2 b
3 c
4 d
Explanation:
D Law of radioactive decay is given as, $\frac{\mathrm{dN}}{\mathrm{dt}}=-\lambda \mathrm{N}$ Where, $\frac{\mathrm{dN}}{\mathrm{dt}}=$ Rate of disintegration So, $\mathrm{R}=-\lambda \mathrm{N} \Rightarrow \frac{\mathrm{R}}{\mathrm{N}}=-\lambda=\mathrm{constant}$ Where, $\lambda=$ decay constant So, the graph $\frac{\mathrm{R}}{\mathrm{N}}$ versus $\mathrm{t}$ is as straight line parallel to $\mathrm{X}$ axis. So, the correct graph shown in option (d).
AP EAMCET (21.04.2019) Shift-I
NUCLEAR PHYSICS
147609
Half - life of a radioactive substance is $\mathbf{1 8}$ minutes. The time interval between its $20 \%$ decay and $\mathbf{8 0} \%$ decay in minutes is
1 6
2 9
3 18
4 36
Explanation:
D Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=18 \mathrm{~min}$ After $n$ half-life the numbers of atom left undecayed is given by- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ For $20 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{20}{100} \mathrm{~N}_{0}=0.8 \mathrm{~N}_{0}$ $0.8 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}}{18}}$ For $80 \%$ decay of radioactive substance, $\mathrm{N}=\mathrm{N}_{0}-\frac{80}{100} \mathrm{~N}_{0}=0.2 \mathrm{~N}_{0}$ $0.2 \mathrm{~N}_{0}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{2}}{18}}$ On dividing equation (i) and (ii), we get- $4=\left(\frac{1}{2}\right)^{\frac{t_{1}-t_{2}}{18}}$ Taking $\log$ on the both sides, we get- $\log (4)=\log \left(\frac{1}{2}\right)^{\frac{\mathrm{t}_{1}-\mathrm{t}_{2}}{18}}$ $\log (2)^{2}=\log (2)^{\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}}$ $2=\frac{\mathrm{t}_{2}-\mathrm{t}_{1}}{18}$ $36=\mathrm{t}_{2}-\mathrm{t}_{1}$ So, $\quad \mathrm{t}_{2}-\mathrm{t}_{1}=36$
AP EAMCET (20.04.2019) Shift-II
NUCLEAR PHYSICS
147610
In a nuclear reactor the activity of a radioactive substance is $2000/ \mathrm{s}$. If the mean life of the products is $\mathbf{5 0}$ minutes, then in the steady power generation, the number of radio nuclides is
1 $12 \times 10^{5}$
2 $60 \times 10^{5}$
3 $90 \times 10^{5}$
4 $15 \times 10^{5}$
Explanation:
B Given, $\frac{\mathrm{dN}}{\mathrm{dt}}=2000 / \mathrm{s}$ Mean life $(\tau)=50 \mathrm{~min}=50 \times 60 \mathrm{sec}$ Since, we know that mean-life of the radioactive substance is inversely proportional to disintegration constant $\lambda$. $\therefore \quad \tau=\frac{1}{\lambda}$ $\lambda=\frac{1}{\tau}=\frac{1}{50 \times 60} \text { per second }$ The rate of decay is proportional to the number of radio nuclides is given as, $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ $2000=\frac{1}{50 \times 60} \times \mathrm{N}$ $\mathrm{N}=2000\times 50 \times 60$ $\mathrm{~N}=60 \times 10^{5}$ Hence, the number of radio nuclides is $60 \times 10^{5}$.
