147594
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\frac{1}{10 \lambda}$
2 $\frac{1}{11 \lambda}$
3 $\frac{11}{10 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
D We know that, The first order decay process, and same no. of initial no. of atoms $\mathrm{X}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda \mathrm{t}}$ $\mathrm{X}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{X_{1}}{X_{2}}=\frac{e^{-10 \lambda t}}{e^{-\lambda t}}$ $\frac{X_{1}}{X_{2}}=e^{-9 \lambda t}$ Determining the time when, $\frac{X_{1}}{X_{2}}=\frac{1}{e}$ $\frac{1}{\mathrm{e}}=\mathrm{e}^{-9 \lambda \mathrm{t}}$ $\mathrm{e}=\mathrm{e}^{9 \lambda}$ $\mathrm{t}=\frac{1}{9 \lambda}$
TS EAMCET 08.05.2019
NUCLEAR PHYSICS
147595
The end product of decay of ${ }_{90} \mathrm{Th}^{232}$ is ${ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$ and $\beta$ particles emitted are respectively.
1 3,3
2 6,4
3 6,0
4 4,6
Explanation:
B Let $\alpha$-particles emitted are $\mathrm{x}$ and $\beta$-particles emitted are y $\left.{ }_{90} \mathrm{Th}^{232} \rightarrow{ }_{82} \mathrm{~Pb}^{208}+\mathrm{x}_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ On comparing atomic number $90=82+2 x-y$ $2 x-y=8$ On comparing mass number $232=208+4 x$ $x=6$ Putting the value of $x$ in equation (i), we get $2 \times 6-y=8$ $y=12-8$ $y=4$ So, number of $\alpha$ and $\beta$ particle are emitted is 6,4 .
Karnataka CET-2019
NUCLEAR PHYSICS
147598
Radioactive element decays to from a stable nuclide, then the rate of decay of reactant is shown by
1 a
2 b
3 c
4 d
Explanation:
C According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{\mathrm{o}} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{e}^{-\lambda \mathrm{t}}$ Rate of decay $\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)$ varies exponentially with time $(\mathrm{t})$.
VITEEE-2019
NUCLEAR PHYSICS
147596
Which one of the following nuclei has shorter mean life?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 Same for all
Explanation:
A The activity of such nuclei is determined by the slope of the supplied curves. In addition, the mean lifetime is inversely proportional to activity as a result, the longer the activity, the shorter the substances mean life. So, nucleus A has the steepest slope nucleus A has the shortest mean life among three.
147594
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\frac{1}{10 \lambda}$
2 $\frac{1}{11 \lambda}$
3 $\frac{11}{10 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
D We know that, The first order decay process, and same no. of initial no. of atoms $\mathrm{X}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda \mathrm{t}}$ $\mathrm{X}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{X_{1}}{X_{2}}=\frac{e^{-10 \lambda t}}{e^{-\lambda t}}$ $\frac{X_{1}}{X_{2}}=e^{-9 \lambda t}$ Determining the time when, $\frac{X_{1}}{X_{2}}=\frac{1}{e}$ $\frac{1}{\mathrm{e}}=\mathrm{e}^{-9 \lambda \mathrm{t}}$ $\mathrm{e}=\mathrm{e}^{9 \lambda}$ $\mathrm{t}=\frac{1}{9 \lambda}$
TS EAMCET 08.05.2019
NUCLEAR PHYSICS
147595
The end product of decay of ${ }_{90} \mathrm{Th}^{232}$ is ${ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$ and $\beta$ particles emitted are respectively.
1 3,3
2 6,4
3 6,0
4 4,6
Explanation:
B Let $\alpha$-particles emitted are $\mathrm{x}$ and $\beta$-particles emitted are y $\left.{ }_{90} \mathrm{Th}^{232} \rightarrow{ }_{82} \mathrm{~Pb}^{208}+\mathrm{x}_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ On comparing atomic number $90=82+2 x-y$ $2 x-y=8$ On comparing mass number $232=208+4 x$ $x=6$ Putting the value of $x$ in equation (i), we get $2 \times 6-y=8$ $y=12-8$ $y=4$ So, number of $\alpha$ and $\beta$ particle are emitted is 6,4 .
Karnataka CET-2019
NUCLEAR PHYSICS
147598
Radioactive element decays to from a stable nuclide, then the rate of decay of reactant is shown by
1 a
2 b
3 c
4 d
Explanation:
C According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{\mathrm{o}} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{e}^{-\lambda \mathrm{t}}$ Rate of decay $\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)$ varies exponentially with time $(\mathrm{t})$.
VITEEE-2019
NUCLEAR PHYSICS
147596
Which one of the following nuclei has shorter mean life?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 Same for all
Explanation:
A The activity of such nuclei is determined by the slope of the supplied curves. In addition, the mean lifetime is inversely proportional to activity as a result, the longer the activity, the shorter the substances mean life. So, nucleus A has the steepest slope nucleus A has the shortest mean life among three.
147594
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\frac{1}{10 \lambda}$
2 $\frac{1}{11 \lambda}$
3 $\frac{11}{10 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
D We know that, The first order decay process, and same no. of initial no. of atoms $\mathrm{X}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda \mathrm{t}}$ $\mathrm{X}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{X_{1}}{X_{2}}=\frac{e^{-10 \lambda t}}{e^{-\lambda t}}$ $\frac{X_{1}}{X_{2}}=e^{-9 \lambda t}$ Determining the time when, $\frac{X_{1}}{X_{2}}=\frac{1}{e}$ $\frac{1}{\mathrm{e}}=\mathrm{e}^{-9 \lambda \mathrm{t}}$ $\mathrm{e}=\mathrm{e}^{9 \lambda}$ $\mathrm{t}=\frac{1}{9 \lambda}$
TS EAMCET 08.05.2019
NUCLEAR PHYSICS
147595
The end product of decay of ${ }_{90} \mathrm{Th}^{232}$ is ${ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$ and $\beta$ particles emitted are respectively.
1 3,3
2 6,4
3 6,0
4 4,6
Explanation:
B Let $\alpha$-particles emitted are $\mathrm{x}$ and $\beta$-particles emitted are y $\left.{ }_{90} \mathrm{Th}^{232} \rightarrow{ }_{82} \mathrm{~Pb}^{208}+\mathrm{x}_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ On comparing atomic number $90=82+2 x-y$ $2 x-y=8$ On comparing mass number $232=208+4 x$ $x=6$ Putting the value of $x$ in equation (i), we get $2 \times 6-y=8$ $y=12-8$ $y=4$ So, number of $\alpha$ and $\beta$ particle are emitted is 6,4 .
Karnataka CET-2019
NUCLEAR PHYSICS
147598
Radioactive element decays to from a stable nuclide, then the rate of decay of reactant is shown by
1 a
2 b
3 c
4 d
Explanation:
C According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{\mathrm{o}} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{e}^{-\lambda \mathrm{t}}$ Rate of decay $\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)$ varies exponentially with time $(\mathrm{t})$.
VITEEE-2019
NUCLEAR PHYSICS
147596
Which one of the following nuclei has shorter mean life?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 Same for all
Explanation:
A The activity of such nuclei is determined by the slope of the supplied curves. In addition, the mean lifetime is inversely proportional to activity as a result, the longer the activity, the shorter the substances mean life. So, nucleus A has the steepest slope nucleus A has the shortest mean life among three.
147594
Two radioactive materials $X_{1}$ and $X_{2}$ have decay constants $10 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_{1}$ to that of $X_{2}$ will be $\frac{1}{e}$ after a time
1 $\frac{1}{10 \lambda}$
2 $\frac{1}{11 \lambda}$
3 $\frac{11}{10 \lambda}$
4 $\frac{1}{9 \lambda}$
Explanation:
D We know that, The first order decay process, and same no. of initial no. of atoms $\mathrm{X}_{1}=\mathrm{N}_{0} \mathrm{e}^{-10 \lambda \mathrm{t}}$ $\mathrm{X}_{2}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{X_{1}}{X_{2}}=\frac{e^{-10 \lambda t}}{e^{-\lambda t}}$ $\frac{X_{1}}{X_{2}}=e^{-9 \lambda t}$ Determining the time when, $\frac{X_{1}}{X_{2}}=\frac{1}{e}$ $\frac{1}{\mathrm{e}}=\mathrm{e}^{-9 \lambda \mathrm{t}}$ $\mathrm{e}=\mathrm{e}^{9 \lambda}$ $\mathrm{t}=\frac{1}{9 \lambda}$
TS EAMCET 08.05.2019
NUCLEAR PHYSICS
147595
The end product of decay of ${ }_{90} \mathrm{Th}^{232}$ is ${ }_{82} \mathrm{~Pb}^{208}$. The number of $\alpha$ and $\beta$ particles emitted are respectively.
1 3,3
2 6,4
3 6,0
4 4,6
Explanation:
B Let $\alpha$-particles emitted are $\mathrm{x}$ and $\beta$-particles emitted are y $\left.{ }_{90} \mathrm{Th}^{232} \rightarrow{ }_{82} \mathrm{~Pb}^{208}+\mathrm{x}_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ On comparing atomic number $90=82+2 x-y$ $2 x-y=8$ On comparing mass number $232=208+4 x$ $x=6$ Putting the value of $x$ in equation (i), we get $2 \times 6-y=8$ $y=12-8$ $y=4$ So, number of $\alpha$ and $\beta$ particle are emitted is 6,4 .
Karnataka CET-2019
NUCLEAR PHYSICS
147598
Radioactive element decays to from a stable nuclide, then the rate of decay of reactant is shown by
1 a
2 b
3 c
4 d
Explanation:
C According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{\mathrm{o}} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{e}^{-\lambda \mathrm{t}}$ Rate of decay $\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)$ varies exponentially with time $(\mathrm{t})$.
VITEEE-2019
NUCLEAR PHYSICS
147596
Which one of the following nuclei has shorter mean life?
1 $\mathrm{A}$
2 B
3 $\mathrm{C}$
4 Same for all
Explanation:
A The activity of such nuclei is determined by the slope of the supplied curves. In addition, the mean lifetime is inversely proportional to activity as a result, the longer the activity, the shorter the substances mean life. So, nucleus A has the steepest slope nucleus A has the shortest mean life among three.
