147851
Radioactive element decays to from the stable nuclide, then the rate of decay of reactant is
1 a
2 b
3 c
4 d
Explanation:
C We know that, radioactive nuclei, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{0} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ Hence, rate of decay $\frac{\mathrm{dN}}{\mathrm{dt}}$ varies exponentially with time.
AIIMS-2012
NUCLEAR PHYSICS
147852
Activity of a radioactive sample decreases to $(1 / 3)^{\text {rd }}$ of its original value in 3 days. Then, in 9 days its activity will become
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A Let the initial activity $\mathrm{N}_{0}$ and time $\mathrm{t}=0$, We know that, Activity of radioactive sample, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ After 3 days, $\mathrm{N}=\frac{\mathrm{N}_{0}}{3}$ $\text { Hence, } \frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\mathrm{e}^{-3 \lambda}=\frac{1}{3}$ After 9 days, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-9 \lambda}$ $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Putting the value of $\mathrm{e}^{3 \lambda}$ in equation (i), we get- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{3}\right)^{3} \Rightarrow \mathrm{N}=\frac{\mathrm{N}_{0}}{27}$
AIIMS-2009
NUCLEAR PHYSICS
147853
The half life of a radioactive substance against $\alpha$-decay is $1.2 \times 10^{7} \mathrm{~s}$. What is the decay rate for $4.0 \times 10^{15}$ atoms of the substance
147854
Starting with a sample of pure ${ }^{66} \mathrm{Cu}, \frac{7}{8}$ of it decays into $\mathrm{Zn}$ in 15 minutes. The corresponding half life is
1 15 minutes
2 10 minutes
3 $7 \frac{1}{2}$ minutes
4 5 minutes
Explanation:
D According to question, $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{7}{8}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, radioactive decay, $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Putting the value, we get- $\frac{1}{8}=\mathrm{e}^{-\lambda \mathrm{t}}$ $8=\mathrm{e}^{\lambda \mathrm{t}}$ Taking natural logarithm both the side we get- $\ln 8=\ln \mathrm{e}^{\lambda \mathrm{t}}$ $3 \ln 2=\lambda \mathrm{t}$ $\lambda=\frac{3 \ln (2)}{\mathrm{t}}$ $\lambda=\frac{3 \times 0.693}{15}$ Half-life period, $T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{3 \times 0.693} \times 15$ $\mathrm{T}_{1 / 2}=5 \mathrm{~min}$
147851
Radioactive element decays to from the stable nuclide, then the rate of decay of reactant is
1 a
2 b
3 c
4 d
Explanation:
C We know that, radioactive nuclei, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{0} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ Hence, rate of decay $\frac{\mathrm{dN}}{\mathrm{dt}}$ varies exponentially with time.
AIIMS-2012
NUCLEAR PHYSICS
147852
Activity of a radioactive sample decreases to $(1 / 3)^{\text {rd }}$ of its original value in 3 days. Then, in 9 days its activity will become
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A Let the initial activity $\mathrm{N}_{0}$ and time $\mathrm{t}=0$, We know that, Activity of radioactive sample, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ After 3 days, $\mathrm{N}=\frac{\mathrm{N}_{0}}{3}$ $\text { Hence, } \frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\mathrm{e}^{-3 \lambda}=\frac{1}{3}$ After 9 days, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-9 \lambda}$ $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Putting the value of $\mathrm{e}^{3 \lambda}$ in equation (i), we get- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{3}\right)^{3} \Rightarrow \mathrm{N}=\frac{\mathrm{N}_{0}}{27}$
AIIMS-2009
NUCLEAR PHYSICS
147853
The half life of a radioactive substance against $\alpha$-decay is $1.2 \times 10^{7} \mathrm{~s}$. What is the decay rate for $4.0 \times 10^{15}$ atoms of the substance
147854
Starting with a sample of pure ${ }^{66} \mathrm{Cu}, \frac{7}{8}$ of it decays into $\mathrm{Zn}$ in 15 minutes. The corresponding half life is
1 15 minutes
2 10 minutes
3 $7 \frac{1}{2}$ minutes
4 5 minutes
Explanation:
D According to question, $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{7}{8}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, radioactive decay, $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Putting the value, we get- $\frac{1}{8}=\mathrm{e}^{-\lambda \mathrm{t}}$ $8=\mathrm{e}^{\lambda \mathrm{t}}$ Taking natural logarithm both the side we get- $\ln 8=\ln \mathrm{e}^{\lambda \mathrm{t}}$ $3 \ln 2=\lambda \mathrm{t}$ $\lambda=\frac{3 \ln (2)}{\mathrm{t}}$ $\lambda=\frac{3 \times 0.693}{15}$ Half-life period, $T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{3 \times 0.693} \times 15$ $\mathrm{T}_{1 / 2}=5 \mathrm{~min}$
147851
Radioactive element decays to from the stable nuclide, then the rate of decay of reactant is
1 a
2 b
3 c
4 d
Explanation:
C We know that, radioactive nuclei, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{0} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ Hence, rate of decay $\frac{\mathrm{dN}}{\mathrm{dt}}$ varies exponentially with time.
AIIMS-2012
NUCLEAR PHYSICS
147852
Activity of a radioactive sample decreases to $(1 / 3)^{\text {rd }}$ of its original value in 3 days. Then, in 9 days its activity will become
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A Let the initial activity $\mathrm{N}_{0}$ and time $\mathrm{t}=0$, We know that, Activity of radioactive sample, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ After 3 days, $\mathrm{N}=\frac{\mathrm{N}_{0}}{3}$ $\text { Hence, } \frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\mathrm{e}^{-3 \lambda}=\frac{1}{3}$ After 9 days, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-9 \lambda}$ $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Putting the value of $\mathrm{e}^{3 \lambda}$ in equation (i), we get- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{3}\right)^{3} \Rightarrow \mathrm{N}=\frac{\mathrm{N}_{0}}{27}$
AIIMS-2009
NUCLEAR PHYSICS
147853
The half life of a radioactive substance against $\alpha$-decay is $1.2 \times 10^{7} \mathrm{~s}$. What is the decay rate for $4.0 \times 10^{15}$ atoms of the substance
147854
Starting with a sample of pure ${ }^{66} \mathrm{Cu}, \frac{7}{8}$ of it decays into $\mathrm{Zn}$ in 15 minutes. The corresponding half life is
1 15 minutes
2 10 minutes
3 $7 \frac{1}{2}$ minutes
4 5 minutes
Explanation:
D According to question, $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{7}{8}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, radioactive decay, $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Putting the value, we get- $\frac{1}{8}=\mathrm{e}^{-\lambda \mathrm{t}}$ $8=\mathrm{e}^{\lambda \mathrm{t}}$ Taking natural logarithm both the side we get- $\ln 8=\ln \mathrm{e}^{\lambda \mathrm{t}}$ $3 \ln 2=\lambda \mathrm{t}$ $\lambda=\frac{3 \ln (2)}{\mathrm{t}}$ $\lambda=\frac{3 \times 0.693}{15}$ Half-life period, $T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{3 \times 0.693} \times 15$ $\mathrm{T}_{1 / 2}=5 \mathrm{~min}$
147851
Radioactive element decays to from the stable nuclide, then the rate of decay of reactant is
1 a
2 b
3 c
4 d
Explanation:
C We know that, radioactive nuclei, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\frac{\mathrm{dN}}{\mathrm{dt}}=-\mathrm{N}_{0} \lambda \mathrm{e}^{-\lambda \mathrm{t}}$ Hence, rate of decay $\frac{\mathrm{dN}}{\mathrm{dt}}$ varies exponentially with time.
AIIMS-2012
NUCLEAR PHYSICS
147852
Activity of a radioactive sample decreases to $(1 / 3)^{\text {rd }}$ of its original value in 3 days. Then, in 9 days its activity will become
1 $(1 / 27)$ of the original value
2 $(1 / 9)$ of the original value
3 $(1 / 18)$ of the original value
4 $(1 / 3)$ of the original value
Explanation:
A Let the initial activity $\mathrm{N}_{0}$ and time $\mathrm{t}=0$, We know that, Activity of radioactive sample, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ After 3 days, $\mathrm{N}=\frac{\mathrm{N}_{0}}{3}$ $\text { Hence, } \frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\mathrm{e}^{-3 \lambda}=\frac{1}{3}$ After 9 days, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-9 \lambda}$ $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Putting the value of $\mathrm{e}^{3 \lambda}$ in equation (i), we get- $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{3}\right)^{3} \Rightarrow \mathrm{N}=\frac{\mathrm{N}_{0}}{27}$
AIIMS-2009
NUCLEAR PHYSICS
147853
The half life of a radioactive substance against $\alpha$-decay is $1.2 \times 10^{7} \mathrm{~s}$. What is the decay rate for $4.0 \times 10^{15}$ atoms of the substance
147854
Starting with a sample of pure ${ }^{66} \mathrm{Cu}, \frac{7}{8}$ of it decays into $\mathrm{Zn}$ in 15 minutes. The corresponding half life is
1 15 minutes
2 10 minutes
3 $7 \frac{1}{2}$ minutes
4 5 minutes
Explanation:
D According to question, $\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $1-\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{7}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=1-\frac{7}{8}=\frac{1}{8}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{8}$ We know that, radioactive decay, $\mathrm{N}=\mathrm{N}_{0}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \mathrm{t}}$ Putting the value, we get- $\frac{1}{8}=\mathrm{e}^{-\lambda \mathrm{t}}$ $8=\mathrm{e}^{\lambda \mathrm{t}}$ Taking natural logarithm both the side we get- $\ln 8=\ln \mathrm{e}^{\lambda \mathrm{t}}$ $3 \ln 2=\lambda \mathrm{t}$ $\lambda=\frac{3 \ln (2)}{\mathrm{t}}$ $\lambda=\frac{3 \times 0.693}{15}$ Half-life period, $T_{1 / 2}=\frac{\ln 2}{\lambda}=\frac{0.693}{3 \times 0.693} \times 15$ $\mathrm{T}_{1 / 2}=5 \mathrm{~min}$
