147811
Two different radioactive elements with halflives ' $T_{1}$ ' and ' $T_{2}$ ' have undecayed atoms ' $N_{1}$ ' and ' $\mathrm{N}_{2}$ ' respectively, present at a given instant. The ratio of their activities at this instant is
A Activity for a radioactive substance is given as $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ So, $\quad \mathrm{R}=\frac{0.693}{\mathrm{~T}} \mathrm{~N} \quad\left(\lambda=\frac{0.693}{\mathrm{~T}}\right)$ Hence, $\mathrm{R}_{1}=\frac{0.693}{\mathrm{~T}_{1}} \mathrm{~N}_{1}$ and $\mathrm{R}_{2}=\frac{0.693}{\mathrm{~T}_{2}} \mathrm{~N}_{2}$ The ratio of their activities, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
MHT-CET 2020
NUCLEAR PHYSICS
147812
If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is .......... days.
1 6
2 4
3 8
4 12
Explanation:
C Initial amount $\left(\mathrm{N}_{\mathrm{o}}\right)=100 \%$ Amount remaining after 16 days $\left(\mathrm{N}_{\mathrm{t}}\right)=25 \%$ $\mathrm{t}=16 \text { days }$ From radioactive decay, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{(-\lambda \mathrm{t})}$ $\lambda=-\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{\mathrm{o}}}\right)}{\mathrm{t}}$ $\lambda=-\frac{\ln (0.25)}{16}=0.086 \text { per day }$ Now, the half-life, $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{\lambda}$ $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{0.086}$ $\mathrm{t}_{1 / 2}=8.06 \text { days } = 8 \text { days }$ Therefore, the half life of the radioactive sample is approximately 8 days.
AP EAMCET (22.09.2020) Shift-II
NUCLEAR PHYSICS
147813
The half-life of radioactive sample is $T$. The fraction of the initial mass of the sample that decays in an interval $T / 2$ is
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 $\frac{(\sqrt{2}-1)}{\sqrt{2}}$
4 $\frac{(\sqrt{2}+1)}{\sqrt{2}}$
Explanation:
A Given that, $\mathrm{t}=\frac{\mathrm{T}}{2}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{t}}{\mathrm{T}}}$ Where, $\mathrm{T}=$ half -life So, $\quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{T} / 2}{\mathrm{~T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{1}{2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\sqrt{2}}$
AP EAMCET (21.09.2020) Shift-II
NUCLEAR PHYSICS
147814
The rate of radioactive disintegration at an instant for a radioactive sample of half life $\mathbf{2 . 2}$ $\times 10^{9} \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is
1 $3.17 \times 10^{20}$
2 $3.17 \times 10^{17}$
3 $3.17 \times 10^{18}$
4 $3.17 \times 10^{19}$
Explanation:
D Given that, Half-life of radioactive sample $\left(\mathrm{T}_{1 / 2}\right)=2.2 \times 10^{9} \mathrm{sec}$. $\mathrm{R}=10^{10} \text { per second }$ We know that, Activity of sample - $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{N}=\frac{\mathrm{R}}{\lambda}$ $\mathrm{N}=\frac{\mathrm{R}}{\frac{0.693}{\mathrm{~T}_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\mathrm{N}=\frac{\mathrm{R} \times \mathrm{T}_{1 / 2}}{0.693}$ $\mathrm{~N}=\frac{10^{10} \times 2.2 \times 10^{9}}{0.693}$ $\mathrm{~N}=\frac{2.2}{0.693} \times 10^{19}$ $\mathrm{~N}=3.17 \times 10^{19} \text { atoms }$
147811
Two different radioactive elements with halflives ' $T_{1}$ ' and ' $T_{2}$ ' have undecayed atoms ' $N_{1}$ ' and ' $\mathrm{N}_{2}$ ' respectively, present at a given instant. The ratio of their activities at this instant is
A Activity for a radioactive substance is given as $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ So, $\quad \mathrm{R}=\frac{0.693}{\mathrm{~T}} \mathrm{~N} \quad\left(\lambda=\frac{0.693}{\mathrm{~T}}\right)$ Hence, $\mathrm{R}_{1}=\frac{0.693}{\mathrm{~T}_{1}} \mathrm{~N}_{1}$ and $\mathrm{R}_{2}=\frac{0.693}{\mathrm{~T}_{2}} \mathrm{~N}_{2}$ The ratio of their activities, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
MHT-CET 2020
NUCLEAR PHYSICS
147812
If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is .......... days.
1 6
2 4
3 8
4 12
Explanation:
C Initial amount $\left(\mathrm{N}_{\mathrm{o}}\right)=100 \%$ Amount remaining after 16 days $\left(\mathrm{N}_{\mathrm{t}}\right)=25 \%$ $\mathrm{t}=16 \text { days }$ From radioactive decay, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{(-\lambda \mathrm{t})}$ $\lambda=-\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{\mathrm{o}}}\right)}{\mathrm{t}}$ $\lambda=-\frac{\ln (0.25)}{16}=0.086 \text { per day }$ Now, the half-life, $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{\lambda}$ $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{0.086}$ $\mathrm{t}_{1 / 2}=8.06 \text { days } = 8 \text { days }$ Therefore, the half life of the radioactive sample is approximately 8 days.
AP EAMCET (22.09.2020) Shift-II
NUCLEAR PHYSICS
147813
The half-life of radioactive sample is $T$. The fraction of the initial mass of the sample that decays in an interval $T / 2$ is
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 $\frac{(\sqrt{2}-1)}{\sqrt{2}}$
4 $\frac{(\sqrt{2}+1)}{\sqrt{2}}$
Explanation:
A Given that, $\mathrm{t}=\frac{\mathrm{T}}{2}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{t}}{\mathrm{T}}}$ Where, $\mathrm{T}=$ half -life So, $\quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{T} / 2}{\mathrm{~T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{1}{2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\sqrt{2}}$
AP EAMCET (21.09.2020) Shift-II
NUCLEAR PHYSICS
147814
The rate of radioactive disintegration at an instant for a radioactive sample of half life $\mathbf{2 . 2}$ $\times 10^{9} \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is
1 $3.17 \times 10^{20}$
2 $3.17 \times 10^{17}$
3 $3.17 \times 10^{18}$
4 $3.17 \times 10^{19}$
Explanation:
D Given that, Half-life of radioactive sample $\left(\mathrm{T}_{1 / 2}\right)=2.2 \times 10^{9} \mathrm{sec}$. $\mathrm{R}=10^{10} \text { per second }$ We know that, Activity of sample - $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{N}=\frac{\mathrm{R}}{\lambda}$ $\mathrm{N}=\frac{\mathrm{R}}{\frac{0.693}{\mathrm{~T}_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\mathrm{N}=\frac{\mathrm{R} \times \mathrm{T}_{1 / 2}}{0.693}$ $\mathrm{~N}=\frac{10^{10} \times 2.2 \times 10^{9}}{0.693}$ $\mathrm{~N}=\frac{2.2}{0.693} \times 10^{19}$ $\mathrm{~N}=3.17 \times 10^{19} \text { atoms }$
147811
Two different radioactive elements with halflives ' $T_{1}$ ' and ' $T_{2}$ ' have undecayed atoms ' $N_{1}$ ' and ' $\mathrm{N}_{2}$ ' respectively, present at a given instant. The ratio of their activities at this instant is
A Activity for a radioactive substance is given as $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ So, $\quad \mathrm{R}=\frac{0.693}{\mathrm{~T}} \mathrm{~N} \quad\left(\lambda=\frac{0.693}{\mathrm{~T}}\right)$ Hence, $\mathrm{R}_{1}=\frac{0.693}{\mathrm{~T}_{1}} \mathrm{~N}_{1}$ and $\mathrm{R}_{2}=\frac{0.693}{\mathrm{~T}_{2}} \mathrm{~N}_{2}$ The ratio of their activities, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
MHT-CET 2020
NUCLEAR PHYSICS
147812
If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is .......... days.
1 6
2 4
3 8
4 12
Explanation:
C Initial amount $\left(\mathrm{N}_{\mathrm{o}}\right)=100 \%$ Amount remaining after 16 days $\left(\mathrm{N}_{\mathrm{t}}\right)=25 \%$ $\mathrm{t}=16 \text { days }$ From radioactive decay, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{(-\lambda \mathrm{t})}$ $\lambda=-\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{\mathrm{o}}}\right)}{\mathrm{t}}$ $\lambda=-\frac{\ln (0.25)}{16}=0.086 \text { per day }$ Now, the half-life, $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{\lambda}$ $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{0.086}$ $\mathrm{t}_{1 / 2}=8.06 \text { days } = 8 \text { days }$ Therefore, the half life of the radioactive sample is approximately 8 days.
AP EAMCET (22.09.2020) Shift-II
NUCLEAR PHYSICS
147813
The half-life of radioactive sample is $T$. The fraction of the initial mass of the sample that decays in an interval $T / 2$ is
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 $\frac{(\sqrt{2}-1)}{\sqrt{2}}$
4 $\frac{(\sqrt{2}+1)}{\sqrt{2}}$
Explanation:
A Given that, $\mathrm{t}=\frac{\mathrm{T}}{2}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{t}}{\mathrm{T}}}$ Where, $\mathrm{T}=$ half -life So, $\quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{T} / 2}{\mathrm{~T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{1}{2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\sqrt{2}}$
AP EAMCET (21.09.2020) Shift-II
NUCLEAR PHYSICS
147814
The rate of radioactive disintegration at an instant for a radioactive sample of half life $\mathbf{2 . 2}$ $\times 10^{9} \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is
1 $3.17 \times 10^{20}$
2 $3.17 \times 10^{17}$
3 $3.17 \times 10^{18}$
4 $3.17 \times 10^{19}$
Explanation:
D Given that, Half-life of radioactive sample $\left(\mathrm{T}_{1 / 2}\right)=2.2 \times 10^{9} \mathrm{sec}$. $\mathrm{R}=10^{10} \text { per second }$ We know that, Activity of sample - $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{N}=\frac{\mathrm{R}}{\lambda}$ $\mathrm{N}=\frac{\mathrm{R}}{\frac{0.693}{\mathrm{~T}_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\mathrm{N}=\frac{\mathrm{R} \times \mathrm{T}_{1 / 2}}{0.693}$ $\mathrm{~N}=\frac{10^{10} \times 2.2 \times 10^{9}}{0.693}$ $\mathrm{~N}=\frac{2.2}{0.693} \times 10^{19}$ $\mathrm{~N}=3.17 \times 10^{19} \text { atoms }$
147811
Two different radioactive elements with halflives ' $T_{1}$ ' and ' $T_{2}$ ' have undecayed atoms ' $N_{1}$ ' and ' $\mathrm{N}_{2}$ ' respectively, present at a given instant. The ratio of their activities at this instant is
A Activity for a radioactive substance is given as $\mathrm{R}=\frac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}$ So, $\quad \mathrm{R}=\frac{0.693}{\mathrm{~T}} \mathrm{~N} \quad\left(\lambda=\frac{0.693}{\mathrm{~T}}\right)$ Hence, $\mathrm{R}_{1}=\frac{0.693}{\mathrm{~T}_{1}} \mathrm{~N}_{1}$ and $\mathrm{R}_{2}=\frac{0.693}{\mathrm{~T}_{2}} \mathrm{~N}_{2}$ The ratio of their activities, $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~T}_{1}} \times \frac{\mathrm{T}_{2}}{\mathrm{~N}_{2}}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{\mathrm{N}_{1}}{\mathrm{~N}_{2}} \times \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
MHT-CET 2020
NUCLEAR PHYSICS
147812
If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is .......... days.
1 6
2 4
3 8
4 12
Explanation:
C Initial amount $\left(\mathrm{N}_{\mathrm{o}}\right)=100 \%$ Amount remaining after 16 days $\left(\mathrm{N}_{\mathrm{t}}\right)=25 \%$ $\mathrm{t}=16 \text { days }$ From radioactive decay, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{(-\lambda \mathrm{t})}$ $\lambda=-\frac{\ln \left(\frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{\mathrm{o}}}\right)}{\mathrm{t}}$ $\lambda=-\frac{\ln (0.25)}{16}=0.086 \text { per day }$ Now, the half-life, $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{\lambda}$ $\mathrm{t}_{1 / 2}=\frac{\ln (2)}{0.086}$ $\mathrm{t}_{1 / 2}=8.06 \text { days } = 8 \text { days }$ Therefore, the half life of the radioactive sample is approximately 8 days.
AP EAMCET (22.09.2020) Shift-II
NUCLEAR PHYSICS
147813
The half-life of radioactive sample is $T$. The fraction of the initial mass of the sample that decays in an interval $T / 2$ is
1 $\frac{1}{\sqrt{2}}$
2 $\sqrt{2}$
3 $\frac{(\sqrt{2}-1)}{\sqrt{2}}$
4 $\frac{(\sqrt{2}+1)}{\sqrt{2}}$
Explanation:
A Given that, $\mathrm{t}=\frac{\mathrm{T}}{2}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{t}}{\mathrm{T}}}$ Where, $\mathrm{T}=$ half -life So, $\quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{\mathrm{T} / 2}{\mathrm{~T}}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left[\frac{1}{2}\right]^{\frac{1}{2}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\sqrt{2}}$
AP EAMCET (21.09.2020) Shift-II
NUCLEAR PHYSICS
147814
The rate of radioactive disintegration at an instant for a radioactive sample of half life $\mathbf{2 . 2}$ $\times 10^{9} \mathrm{~s}$ is $10^{10} \mathrm{~s}^{-1}$. The number of radioactive atoms in that sample at that instant is
1 $3.17 \times 10^{20}$
2 $3.17 \times 10^{17}$
3 $3.17 \times 10^{18}$
4 $3.17 \times 10^{19}$
Explanation:
D Given that, Half-life of radioactive sample $\left(\mathrm{T}_{1 / 2}\right)=2.2 \times 10^{9} \mathrm{sec}$. $\mathrm{R}=10^{10} \text { per second }$ We know that, Activity of sample - $\mathrm{R}=\lambda \mathrm{N}$ $\mathrm{N}=\frac{\mathrm{R}}{\lambda}$ $\mathrm{N}=\frac{\mathrm{R}}{\frac{0.693}{\mathrm{~T}_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{\mathrm{~T}_{1 / 2}}\right)$ $\mathrm{N}=\frac{\mathrm{R} \times \mathrm{T}_{1 / 2}}{0.693}$ $\mathrm{~N}=\frac{10^{10} \times 2.2 \times 10^{9}}{0.693}$ $\mathrm{~N}=\frac{2.2}{0.693} \times 10^{19}$ $\mathrm{~N}=3.17 \times 10^{19} \text { atoms }$
