1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C Nuclear stability depends upon the ratio of neutron to proton. If the $\frac{\mathrm{n}}{\mathrm{p}}$ ratio is more than the critical value, then a neutron gets converted into a proton forming a $\beta$ particle. $\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{v}$ The $\beta$-particle (e) is emitted from the nucleus in some radioactive transformation. So, electron do not exist in the nucleus but result in some nuclear transformation. So, assertion is true but reason is false.
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147601
$99 \%$ of a radioactive element will decay between
1 6 and 7 half lives
2 7 and 8 half lives
3 8 and 9 half lives
4 9 half lives
Explanation:
A When $99 \%$ material decays, the quantity that is left behind is $1 \%$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{100}$ $\because \quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $(2)^{\mathrm{n}}=100$ $2^{\mathrm{n}}=100 \text { we find that } 2^{6}=64 \text { and } 2^{7}=128$ So, $\mathrm{n}$ lies between 6 and 7 half lives.
AIIMS-26.05.2019 (E) Shift-2
NUCLEAR PHYSICS
147602
A sample has half life of $10^{33}$ year. If initial number of nuclei of the sample is $26 \times 10^{24}$. Then find out the number of nuclei decayed in 1 year.
1 $1.82 \times 10^{-9}$
2 $182 \times 10^{-9}$
3 $18.2 \times 10^{-9}$
4 $1820 \times 10^{-9}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=10^{33}$ year Initial number of nuclei $\left(\mathrm{N}_{0}\right)=26 \times 10^{24}$ Total time $(\mathrm{t})=1$ year As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{10^{33}}=0.693 \times 10^{-33} / \text { years }$ According to radioactive decay law - $\left|\frac{-\mathrm{dN}}{\mathrm{dt}}\right|=\lambda \mathrm{N}_{0}$ $\mathrm{dN}=\lambda \mathrm{N}_{0} \mathrm{dt}$ $\mathrm{dN}=0.693 \times 10^{-33} \times 26 \times 10^{24} \times 1$ $\mathrm{dN}=18.2 \times 10^{-9}$
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147603
If half life an element is 69.3 hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=\mathbf{5 0} \mu \mathrm{Ci}$
1 $1 \%$
2 $2 \%$
3 $3 \%$
4 $4 \%$
Explanation:
A Given, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=69.3 \text { hours }$ $\qquad \mathrm{t}_{\mathrm{A}}=10 \mathrm{~h}, \mathrm{t}_{\mathrm{B}}=11 \mathrm{~h}$ $\quad \mathrm{~N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}$ $\text { Activity of nuclei at } \mathrm{t}=11 \text { hours }$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}$ $\quad \% \text { decay }=\left(\frac{\mathrm{N}_{\mathrm{A}}-\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}\right) \times 100$ Putting the value of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ from equation (i) and (ii) in equation (iii), we get - $=\left(\frac{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}}{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / \mathrm{T}_{1 / 2}}}\right) \times 100 \quad\left(\therefore \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / 69.3}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100$ $\%$ decay $=1 \%$ So, the percentage of decay is 1 .
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C Nuclear stability depends upon the ratio of neutron to proton. If the $\frac{\mathrm{n}}{\mathrm{p}}$ ratio is more than the critical value, then a neutron gets converted into a proton forming a $\beta$ particle. $\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{v}$ The $\beta$-particle (e) is emitted from the nucleus in some radioactive transformation. So, electron do not exist in the nucleus but result in some nuclear transformation. So, assertion is true but reason is false.
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147601
$99 \%$ of a radioactive element will decay between
1 6 and 7 half lives
2 7 and 8 half lives
3 8 and 9 half lives
4 9 half lives
Explanation:
A When $99 \%$ material decays, the quantity that is left behind is $1 \%$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{100}$ $\because \quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $(2)^{\mathrm{n}}=100$ $2^{\mathrm{n}}=100 \text { we find that } 2^{6}=64 \text { and } 2^{7}=128$ So, $\mathrm{n}$ lies between 6 and 7 half lives.
AIIMS-26.05.2019 (E) Shift-2
NUCLEAR PHYSICS
147602
A sample has half life of $10^{33}$ year. If initial number of nuclei of the sample is $26 \times 10^{24}$. Then find out the number of nuclei decayed in 1 year.
1 $1.82 \times 10^{-9}$
2 $182 \times 10^{-9}$
3 $18.2 \times 10^{-9}$
4 $1820 \times 10^{-9}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=10^{33}$ year Initial number of nuclei $\left(\mathrm{N}_{0}\right)=26 \times 10^{24}$ Total time $(\mathrm{t})=1$ year As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{10^{33}}=0.693 \times 10^{-33} / \text { years }$ According to radioactive decay law - $\left|\frac{-\mathrm{dN}}{\mathrm{dt}}\right|=\lambda \mathrm{N}_{0}$ $\mathrm{dN}=\lambda \mathrm{N}_{0} \mathrm{dt}$ $\mathrm{dN}=0.693 \times 10^{-33} \times 26 \times 10^{24} \times 1$ $\mathrm{dN}=18.2 \times 10^{-9}$
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147603
If half life an element is 69.3 hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=\mathbf{5 0} \mu \mathrm{Ci}$
1 $1 \%$
2 $2 \%$
3 $3 \%$
4 $4 \%$
Explanation:
A Given, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=69.3 \text { hours }$ $\qquad \mathrm{t}_{\mathrm{A}}=10 \mathrm{~h}, \mathrm{t}_{\mathrm{B}}=11 \mathrm{~h}$ $\quad \mathrm{~N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}$ $\text { Activity of nuclei at } \mathrm{t}=11 \text { hours }$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}$ $\quad \% \text { decay }=\left(\frac{\mathrm{N}_{\mathrm{A}}-\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}\right) \times 100$ Putting the value of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ from equation (i) and (ii) in equation (iii), we get - $=\left(\frac{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}}{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / \mathrm{T}_{1 / 2}}}\right) \times 100 \quad\left(\therefore \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / 69.3}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100$ $\%$ decay $=1 \%$ So, the percentage of decay is 1 .
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C Nuclear stability depends upon the ratio of neutron to proton. If the $\frac{\mathrm{n}}{\mathrm{p}}$ ratio is more than the critical value, then a neutron gets converted into a proton forming a $\beta$ particle. $\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{v}$ The $\beta$-particle (e) is emitted from the nucleus in some radioactive transformation. So, electron do not exist in the nucleus but result in some nuclear transformation. So, assertion is true but reason is false.
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147601
$99 \%$ of a radioactive element will decay between
1 6 and 7 half lives
2 7 and 8 half lives
3 8 and 9 half lives
4 9 half lives
Explanation:
A When $99 \%$ material decays, the quantity that is left behind is $1 \%$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{100}$ $\because \quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $(2)^{\mathrm{n}}=100$ $2^{\mathrm{n}}=100 \text { we find that } 2^{6}=64 \text { and } 2^{7}=128$ So, $\mathrm{n}$ lies between 6 and 7 half lives.
AIIMS-26.05.2019 (E) Shift-2
NUCLEAR PHYSICS
147602
A sample has half life of $10^{33}$ year. If initial number of nuclei of the sample is $26 \times 10^{24}$. Then find out the number of nuclei decayed in 1 year.
1 $1.82 \times 10^{-9}$
2 $182 \times 10^{-9}$
3 $18.2 \times 10^{-9}$
4 $1820 \times 10^{-9}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=10^{33}$ year Initial number of nuclei $\left(\mathrm{N}_{0}\right)=26 \times 10^{24}$ Total time $(\mathrm{t})=1$ year As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{10^{33}}=0.693 \times 10^{-33} / \text { years }$ According to radioactive decay law - $\left|\frac{-\mathrm{dN}}{\mathrm{dt}}\right|=\lambda \mathrm{N}_{0}$ $\mathrm{dN}=\lambda \mathrm{N}_{0} \mathrm{dt}$ $\mathrm{dN}=0.693 \times 10^{-33} \times 26 \times 10^{24} \times 1$ $\mathrm{dN}=18.2 \times 10^{-9}$
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147603
If half life an element is 69.3 hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=\mathbf{5 0} \mu \mathrm{Ci}$
1 $1 \%$
2 $2 \%$
3 $3 \%$
4 $4 \%$
Explanation:
A Given, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=69.3 \text { hours }$ $\qquad \mathrm{t}_{\mathrm{A}}=10 \mathrm{~h}, \mathrm{t}_{\mathrm{B}}=11 \mathrm{~h}$ $\quad \mathrm{~N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}$ $\text { Activity of nuclei at } \mathrm{t}=11 \text { hours }$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}$ $\quad \% \text { decay }=\left(\frac{\mathrm{N}_{\mathrm{A}}-\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}\right) \times 100$ Putting the value of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ from equation (i) and (ii) in equation (iii), we get - $=\left(\frac{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}}{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / \mathrm{T}_{1 / 2}}}\right) \times 100 \quad\left(\therefore \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / 69.3}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100$ $\%$ decay $=1 \%$ So, the percentage of decay is 1 .
1 If both assertion and reason are true and reason is the correct explanation of assertion.
2 If both assertion and reason are true but reason is not the correct explanation of assertion.
3 If assertion is true but reason is false.
4 If both assertion and reason are false.
Explanation:
C Nuclear stability depends upon the ratio of neutron to proton. If the $\frac{\mathrm{n}}{\mathrm{p}}$ ratio is more than the critical value, then a neutron gets converted into a proton forming a $\beta$ particle. $\mathrm{n} \rightarrow \mathrm{p}+\mathrm{e}^{-}+\bar{v}$ The $\beta$-particle (e) is emitted from the nucleus in some radioactive transformation. So, electron do not exist in the nucleus but result in some nuclear transformation. So, assertion is true but reason is false.
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147601
$99 \%$ of a radioactive element will decay between
1 6 and 7 half lives
2 7 and 8 half lives
3 8 and 9 half lives
4 9 half lives
Explanation:
A When $99 \%$ material decays, the quantity that is left behind is $1 \%$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{100}$ $\because \quad \frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{1}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $(2)^{\mathrm{n}}=100$ $2^{\mathrm{n}}=100 \text { we find that } 2^{6}=64 \text { and } 2^{7}=128$ So, $\mathrm{n}$ lies between 6 and 7 half lives.
AIIMS-26.05.2019 (E) Shift-2
NUCLEAR PHYSICS
147602
A sample has half life of $10^{33}$ year. If initial number of nuclei of the sample is $26 \times 10^{24}$. Then find out the number of nuclei decayed in 1 year.
1 $1.82 \times 10^{-9}$
2 $182 \times 10^{-9}$
3 $18.2 \times 10^{-9}$
4 $1820 \times 10^{-9}$
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=10^{33}$ year Initial number of nuclei $\left(\mathrm{N}_{0}\right)=26 \times 10^{24}$ Total time $(\mathrm{t})=1$ year As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{10^{33}}=0.693 \times 10^{-33} / \text { years }$ According to radioactive decay law - $\left|\frac{-\mathrm{dN}}{\mathrm{dt}}\right|=\lambda \mathrm{N}_{0}$ $\mathrm{dN}=\lambda \mathrm{N}_{0} \mathrm{dt}$ $\mathrm{dN}=0.693 \times 10^{-33} \times 26 \times 10^{24} \times 1$ $\mathrm{dN}=18.2 \times 10^{-9}$
AIIMS-26.05.2019(E) Shift-2
NUCLEAR PHYSICS
147603
If half life an element is 69.3 hours then how much of its percent will decay in $10^{\text {th }}$ to $11^{\text {th }}$ hours. Initial activity $=\mathbf{5 0} \mu \mathrm{Ci}$
1 $1 \%$
2 $2 \%$
3 $3 \%$
4 $4 \%$
Explanation:
A Given, $\text { Half life }\left(\mathrm{T}_{1 / 2}\right)=69.3 \text { hours }$ $\qquad \mathrm{t}_{\mathrm{A}}=10 \mathrm{~h}, \mathrm{t}_{\mathrm{B}}=11 \mathrm{~h}$ $\quad \mathrm{~N}_{\mathrm{A}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}$ $\text { Activity of nuclei at } \mathrm{t}=11 \text { hours }$ $\mathrm{N}_{\mathrm{B}}=\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}$ $\quad \% \text { decay }=\left(\frac{\mathrm{N}_{\mathrm{A}}-\mathrm{N}_{\mathrm{B}}}{\mathrm{N}_{\mathrm{A}}}\right) \times 100$ Putting the value of $\mathrm{N}_{\mathrm{A}}$ and $\mathrm{N}_{\mathrm{B}}$ from equation (i) and (ii) in equation (iii), we get - $=\left(\frac{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}-\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-11 \lambda}}{\mathrm{N}_{\mathrm{o}} \mathrm{e}^{-10 \lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\lambda}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / \mathrm{T}_{1 / 2}}}\right) \times 100 \quad\left(\therefore \lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}\right)$ $=\left(1-\frac{1}{\mathrm{e}^{\ln 2 / 69.3}}\right) \times 100$ $=\left(1-\frac{1}{\mathrm{e}^{0.01}}\right) \times 100$ $\%$ decay $=1 \%$ So, the percentage of decay is 1 .
