147616
The half life of a radioactive material is 24 hours. The time taken for the radioactive material to decay to $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial amount is
1 72 hours
2 24 hours
3 96 hours
4 48 hours
Explanation:
A Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=24$ hours $\mathrm{N}=\frac{\mathrm{N}_{0}}{8}$ We know that, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{24}=0.02887$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{N}_{0}}{8}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=8=2^{3}$ Taking $\log$ on both side $\lambda \times \mathrm{t}=3 \ln 2$ $\mathrm{t}=\frac{3 \times 0.693}{0.02887}$ $\mathrm{t}=72.01 \approx 72 \text { hours }$
COMEDK 2018
NUCLEAR PHYSICS
147620
$1 \mathrm{mg}$ radium has $2.68 \times 10^{8}$ atoms. Its half-life is 1620 years. How many radium atoms will disintegrate from $1 \mathrm{mg}$ of pure radium in 3240 years?
1 $2.01 \times 10^{9}$
2 $2.01 \times 10^{8}$
3 $1.01 \times 10^{9}$
4 $1.01 \times 10^{8}$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1620$ years Decay time $(\mathrm{t})=3240$ years Number of half lives $(n)=\frac{\text { decay time }(t)}{\text { half life }\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{3240}{1620}=2$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{\mathrm{m}}{\mathrm{m}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ Mass of radium left after 2 half lives is- $\mathrm{m}=\mathrm{m}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=1 \times\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=0.25 \mathrm{mg}$ Mass of radium disintegrated $=1-0.25=0.75 \mathrm{mg}$ Number of radium atoms disintegrated $=0.75 \times 2.68 \times 10^{8}$ $=2.01 \times 10^{8}$
JCECE-2018
NUCLEAR PHYSICS
147621
The number of half lives elapsed before $93.75 \%$ of a radioactive sample has decayed is
1 6
2 4
3 2
4 8
Explanation:
B Since, we know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ According to the question, $\mathrm{N}=(100-93.75) \times \frac{\mathrm{N}_{0}}{100}=\frac{6.25}{100} \times \mathrm{N}_{0}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{6.25}{100}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{5}{10}\right)^{4}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{4}$ $\therefore \quad\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ So, $\quad \mathrm{n}=4$.
AP EAMCET (23.04.2018) Shift-2
NUCLEAR PHYSICS
147622
The activity of a radioactive element decreases in 10 years to $1 / 5$ of initial activity $A_{0}$. After further next 10 years, its activity will be
1 $\frac{\mathrm{A}_{0}}{4}$
2 $\frac{\mathrm{A}_{0}}{10}$
3 $\frac{\mathrm{A}_{0}}{15}$
4 $\frac{\mathrm{A}_{0}}{25}$
Explanation:
D Given, Time $(\mathrm{t})=10$ years Remaining activity of radioactive element $(A)=\frac{A_{0}}{5}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-\lambda t}$ Taking logarithm on both side - $\ln \left(\frac{\mathrm{A}}{\mathrm{A}_{0}}\right)=-\lambda \mathrm{t}$ $\ln \left(\frac{\frac{1}{5} \mathrm{~A}_{0}}{\mathrm{~A}_{0}}\right)=-10 \lambda$ $\ln 5=10 \lambda$ $\therefore \quad \lambda=\frac{\ln 5}{10}$ Now, time $(\mathrm{t})=20$ years $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-20 \times \frac{\ln 5}{10}}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{\ln 5^{-2}}$ $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{\ln 1 / 25}$ So, $\quad \mathrm{A}=\frac{\mathrm{A}_{0}}{25}$
147616
The half life of a radioactive material is 24 hours. The time taken for the radioactive material to decay to $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial amount is
1 72 hours
2 24 hours
3 96 hours
4 48 hours
Explanation:
A Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=24$ hours $\mathrm{N}=\frac{\mathrm{N}_{0}}{8}$ We know that, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{24}=0.02887$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{N}_{0}}{8}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=8=2^{3}$ Taking $\log$ on both side $\lambda \times \mathrm{t}=3 \ln 2$ $\mathrm{t}=\frac{3 \times 0.693}{0.02887}$ $\mathrm{t}=72.01 \approx 72 \text { hours }$
COMEDK 2018
NUCLEAR PHYSICS
147620
$1 \mathrm{mg}$ radium has $2.68 \times 10^{8}$ atoms. Its half-life is 1620 years. How many radium atoms will disintegrate from $1 \mathrm{mg}$ of pure radium in 3240 years?
1 $2.01 \times 10^{9}$
2 $2.01 \times 10^{8}$
3 $1.01 \times 10^{9}$
4 $1.01 \times 10^{8}$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1620$ years Decay time $(\mathrm{t})=3240$ years Number of half lives $(n)=\frac{\text { decay time }(t)}{\text { half life }\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{3240}{1620}=2$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{\mathrm{m}}{\mathrm{m}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ Mass of radium left after 2 half lives is- $\mathrm{m}=\mathrm{m}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=1 \times\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=0.25 \mathrm{mg}$ Mass of radium disintegrated $=1-0.25=0.75 \mathrm{mg}$ Number of radium atoms disintegrated $=0.75 \times 2.68 \times 10^{8}$ $=2.01 \times 10^{8}$
JCECE-2018
NUCLEAR PHYSICS
147621
The number of half lives elapsed before $93.75 \%$ of a radioactive sample has decayed is
1 6
2 4
3 2
4 8
Explanation:
B Since, we know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ According to the question, $\mathrm{N}=(100-93.75) \times \frac{\mathrm{N}_{0}}{100}=\frac{6.25}{100} \times \mathrm{N}_{0}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{6.25}{100}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{5}{10}\right)^{4}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{4}$ $\therefore \quad\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ So, $\quad \mathrm{n}=4$.
AP EAMCET (23.04.2018) Shift-2
NUCLEAR PHYSICS
147622
The activity of a radioactive element decreases in 10 years to $1 / 5$ of initial activity $A_{0}$. After further next 10 years, its activity will be
1 $\frac{\mathrm{A}_{0}}{4}$
2 $\frac{\mathrm{A}_{0}}{10}$
3 $\frac{\mathrm{A}_{0}}{15}$
4 $\frac{\mathrm{A}_{0}}{25}$
Explanation:
D Given, Time $(\mathrm{t})=10$ years Remaining activity of radioactive element $(A)=\frac{A_{0}}{5}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-\lambda t}$ Taking logarithm on both side - $\ln \left(\frac{\mathrm{A}}{\mathrm{A}_{0}}\right)=-\lambda \mathrm{t}$ $\ln \left(\frac{\frac{1}{5} \mathrm{~A}_{0}}{\mathrm{~A}_{0}}\right)=-10 \lambda$ $\ln 5=10 \lambda$ $\therefore \quad \lambda=\frac{\ln 5}{10}$ Now, time $(\mathrm{t})=20$ years $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-20 \times \frac{\ln 5}{10}}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{\ln 5^{-2}}$ $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{\ln 1 / 25}$ So, $\quad \mathrm{A}=\frac{\mathrm{A}_{0}}{25}$
147616
The half life of a radioactive material is 24 hours. The time taken for the radioactive material to decay to $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial amount is
1 72 hours
2 24 hours
3 96 hours
4 48 hours
Explanation:
A Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=24$ hours $\mathrm{N}=\frac{\mathrm{N}_{0}}{8}$ We know that, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{24}=0.02887$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{N}_{0}}{8}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=8=2^{3}$ Taking $\log$ on both side $\lambda \times \mathrm{t}=3 \ln 2$ $\mathrm{t}=\frac{3 \times 0.693}{0.02887}$ $\mathrm{t}=72.01 \approx 72 \text { hours }$
COMEDK 2018
NUCLEAR PHYSICS
147620
$1 \mathrm{mg}$ radium has $2.68 \times 10^{8}$ atoms. Its half-life is 1620 years. How many radium atoms will disintegrate from $1 \mathrm{mg}$ of pure radium in 3240 years?
1 $2.01 \times 10^{9}$
2 $2.01 \times 10^{8}$
3 $1.01 \times 10^{9}$
4 $1.01 \times 10^{8}$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1620$ years Decay time $(\mathrm{t})=3240$ years Number of half lives $(n)=\frac{\text { decay time }(t)}{\text { half life }\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{3240}{1620}=2$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{\mathrm{m}}{\mathrm{m}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ Mass of radium left after 2 half lives is- $\mathrm{m}=\mathrm{m}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=1 \times\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=0.25 \mathrm{mg}$ Mass of radium disintegrated $=1-0.25=0.75 \mathrm{mg}$ Number of radium atoms disintegrated $=0.75 \times 2.68 \times 10^{8}$ $=2.01 \times 10^{8}$
JCECE-2018
NUCLEAR PHYSICS
147621
The number of half lives elapsed before $93.75 \%$ of a radioactive sample has decayed is
1 6
2 4
3 2
4 8
Explanation:
B Since, we know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ According to the question, $\mathrm{N}=(100-93.75) \times \frac{\mathrm{N}_{0}}{100}=\frac{6.25}{100} \times \mathrm{N}_{0}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{6.25}{100}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{5}{10}\right)^{4}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{4}$ $\therefore \quad\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ So, $\quad \mathrm{n}=4$.
AP EAMCET (23.04.2018) Shift-2
NUCLEAR PHYSICS
147622
The activity of a radioactive element decreases in 10 years to $1 / 5$ of initial activity $A_{0}$. After further next 10 years, its activity will be
1 $\frac{\mathrm{A}_{0}}{4}$
2 $\frac{\mathrm{A}_{0}}{10}$
3 $\frac{\mathrm{A}_{0}}{15}$
4 $\frac{\mathrm{A}_{0}}{25}$
Explanation:
D Given, Time $(\mathrm{t})=10$ years Remaining activity of radioactive element $(A)=\frac{A_{0}}{5}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-\lambda t}$ Taking logarithm on both side - $\ln \left(\frac{\mathrm{A}}{\mathrm{A}_{0}}\right)=-\lambda \mathrm{t}$ $\ln \left(\frac{\frac{1}{5} \mathrm{~A}_{0}}{\mathrm{~A}_{0}}\right)=-10 \lambda$ $\ln 5=10 \lambda$ $\therefore \quad \lambda=\frac{\ln 5}{10}$ Now, time $(\mathrm{t})=20$ years $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-20 \times \frac{\ln 5}{10}}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{\ln 5^{-2}}$ $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{\ln 1 / 25}$ So, $\quad \mathrm{A}=\frac{\mathrm{A}_{0}}{25}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147616
The half life of a radioactive material is 24 hours. The time taken for the radioactive material to decay to $\left(\frac{1}{8}\right)^{\text {th }}$ of its initial amount is
1 72 hours
2 24 hours
3 96 hours
4 48 hours
Explanation:
A Given, Half-life $\left(\mathrm{T}_{1 / 2}\right)=24$ hours $\mathrm{N}=\frac{\mathrm{N}_{0}}{8}$ We know that, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{24}=0.02887$ According to radioactive decay law, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{N}_{0}}{8}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{e}^{\lambda \mathrm{t}}=8=2^{3}$ Taking $\log$ on both side $\lambda \times \mathrm{t}=3 \ln 2$ $\mathrm{t}=\frac{3 \times 0.693}{0.02887}$ $\mathrm{t}=72.01 \approx 72 \text { hours }$
COMEDK 2018
NUCLEAR PHYSICS
147620
$1 \mathrm{mg}$ radium has $2.68 \times 10^{8}$ atoms. Its half-life is 1620 years. How many radium atoms will disintegrate from $1 \mathrm{mg}$ of pure radium in 3240 years?
1 $2.01 \times 10^{9}$
2 $2.01 \times 10^{8}$
3 $1.01 \times 10^{9}$
4 $1.01 \times 10^{8}$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=1620$ years Decay time $(\mathrm{t})=3240$ years Number of half lives $(n)=\frac{\text { decay time }(t)}{\text { half life }\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{3240}{1620}=2$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{\mathrm{m}}{\mathrm{m}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ Mass of radium left after 2 half lives is- $\mathrm{m}=\mathrm{m}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}=1 \times\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=0.25 \mathrm{mg}$ Mass of radium disintegrated $=1-0.25=0.75 \mathrm{mg}$ Number of radium atoms disintegrated $=0.75 \times 2.68 \times 10^{8}$ $=2.01 \times 10^{8}$
JCECE-2018
NUCLEAR PHYSICS
147621
The number of half lives elapsed before $93.75 \%$ of a radioactive sample has decayed is
1 6
2 4
3 2
4 8
Explanation:
B Since, we know that, $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ According to the question, $\mathrm{N}=(100-93.75) \times \frac{\mathrm{N}_{0}}{100}=\frac{6.25}{100} \times \mathrm{N}_{0}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{6.25}{100}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{5}{10}\right)^{4}$ $\left(\frac{1}{2}\right)^{\mathrm{n}}=\left(\frac{1}{2}\right)^{4}$ $\therefore \quad\left(\frac{1}{2}\right)^{\mathrm{n}}=\frac{625}{10000}$ So, $\quad \mathrm{n}=4$.
AP EAMCET (23.04.2018) Shift-2
NUCLEAR PHYSICS
147622
The activity of a radioactive element decreases in 10 years to $1 / 5$ of initial activity $A_{0}$. After further next 10 years, its activity will be
1 $\frac{\mathrm{A}_{0}}{4}$
2 $\frac{\mathrm{A}_{0}}{10}$
3 $\frac{\mathrm{A}_{0}}{15}$
4 $\frac{\mathrm{A}_{0}}{25}$
Explanation:
D Given, Time $(\mathrm{t})=10$ years Remaining activity of radioactive element $(A)=\frac{A_{0}}{5}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-\lambda t}$ Taking logarithm on both side - $\ln \left(\frac{\mathrm{A}}{\mathrm{A}_{0}}\right)=-\lambda \mathrm{t}$ $\ln \left(\frac{\frac{1}{5} \mathrm{~A}_{0}}{\mathrm{~A}_{0}}\right)=-10 \lambda$ $\ln 5=10 \lambda$ $\therefore \quad \lambda=\frac{\ln 5}{10}$ Now, time $(\mathrm{t})=20$ years $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{-20 \times \frac{\ln 5}{10}}$ $\frac{\mathrm{A}}{\mathrm{A}_{0}}=\mathrm{e}^{\ln 5^{-2}}$ $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{\ln 1 / 25}$ So, $\quad \mathrm{A}=\frac{\mathrm{A}_{0}}{25}$
