147743
Three fourths of the active nuclei present in a radioactive sample decay in $3 / 4 \mathrm{~s}$. The half life of the sample is :
1 $\frac{3}{8} \mathrm{~s}$
2 $\frac{3}{4} \mathrm{~s}$
3 $\frac{1}{2} \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
A Given, Time $(\mathrm{t})=\frac{3}{4} \mathrm{~s}$ After radioactive sample decay remaining is $\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=2$ $\mathrm{T}_{1 / 2}=\frac{\mathrm{t}}{2}=\frac{3}{4 \times 2}=\frac{3}{8} \mathrm{~s}$
Karnataka CET-2001
NUCLEAR PHYSICS
147745
The fraction of a sample of radioactive nuclei that remains undecayed in one mean life is
1 $\frac{1}{\mathrm{e}}$
2 $1-\frac{1}{\mathrm{e}}$
3 $\frac{1}{\mathrm{e}^{2}}$
4 $1-\frac{1}{\mathrm{e}^{2}}$
Explanation:
A According to equation of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Mean life time $(\mathrm{t})=\frac{1}{\lambda}$ Substituting value of mean life in equation (i), we get $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}$
J and K CET- 2010
NUCLEAR PHYSICS
147746
The activity of $1 \mathrm{mg}$ sample of ${ }_{37}^{90} \mathrm{Sr}$ whose half life is 28 years is (Given that Avogadro's number is $\mathbf{6 . 0 2} \times \mathbf{1 0}^{\mathbf{2 3}}$ )
147749
A radioactive substance decays to $(1 / 16)^{\text {th }}$ of its initial activity in 40 days. The half life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given, $\mathrm{t}=40 \text { days, } \frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{4}=\frac{40}{4}=10 \text { days }$
UP CPMT-2014
NUCLEAR PHYSICS
147750
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number of the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \alpha, 4 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D $\mathrm{Z}=90, \mathrm{Th}^{\mathrm{A}=228} \rightarrow \mathrm{Z}_{\mathrm{Z}^{\prime}=83} \mathrm{Bi}^{\mathrm{A}^{\prime}=212}$ Number of $\alpha$-particles emitted $\mathrm{n}_{\alpha}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{4}=\frac{228-212}{4}=4$ Number of $\beta$-particle emitted $\mathrm{n}_{\beta}=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ So, $4 \alpha$ and $1 \beta$ particles are emitted.
147743
Three fourths of the active nuclei present in a radioactive sample decay in $3 / 4 \mathrm{~s}$. The half life of the sample is :
1 $\frac{3}{8} \mathrm{~s}$
2 $\frac{3}{4} \mathrm{~s}$
3 $\frac{1}{2} \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
A Given, Time $(\mathrm{t})=\frac{3}{4} \mathrm{~s}$ After radioactive sample decay remaining is $\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=2$ $\mathrm{T}_{1 / 2}=\frac{\mathrm{t}}{2}=\frac{3}{4 \times 2}=\frac{3}{8} \mathrm{~s}$
Karnataka CET-2001
NUCLEAR PHYSICS
147745
The fraction of a sample of radioactive nuclei that remains undecayed in one mean life is
1 $\frac{1}{\mathrm{e}}$
2 $1-\frac{1}{\mathrm{e}}$
3 $\frac{1}{\mathrm{e}^{2}}$
4 $1-\frac{1}{\mathrm{e}^{2}}$
Explanation:
A According to equation of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Mean life time $(\mathrm{t})=\frac{1}{\lambda}$ Substituting value of mean life in equation (i), we get $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}$
J and K CET- 2010
NUCLEAR PHYSICS
147746
The activity of $1 \mathrm{mg}$ sample of ${ }_{37}^{90} \mathrm{Sr}$ whose half life is 28 years is (Given that Avogadro's number is $\mathbf{6 . 0 2} \times \mathbf{1 0}^{\mathbf{2 3}}$ )
147749
A radioactive substance decays to $(1 / 16)^{\text {th }}$ of its initial activity in 40 days. The half life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given, $\mathrm{t}=40 \text { days, } \frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{4}=\frac{40}{4}=10 \text { days }$
UP CPMT-2014
NUCLEAR PHYSICS
147750
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number of the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \alpha, 4 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D $\mathrm{Z}=90, \mathrm{Th}^{\mathrm{A}=228} \rightarrow \mathrm{Z}_{\mathrm{Z}^{\prime}=83} \mathrm{Bi}^{\mathrm{A}^{\prime}=212}$ Number of $\alpha$-particles emitted $\mathrm{n}_{\alpha}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{4}=\frac{228-212}{4}=4$ Number of $\beta$-particle emitted $\mathrm{n}_{\beta}=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ So, $4 \alpha$ and $1 \beta$ particles are emitted.
147743
Three fourths of the active nuclei present in a radioactive sample decay in $3 / 4 \mathrm{~s}$. The half life of the sample is :
1 $\frac{3}{8} \mathrm{~s}$
2 $\frac{3}{4} \mathrm{~s}$
3 $\frac{1}{2} \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
A Given, Time $(\mathrm{t})=\frac{3}{4} \mathrm{~s}$ After radioactive sample decay remaining is $\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=2$ $\mathrm{T}_{1 / 2}=\frac{\mathrm{t}}{2}=\frac{3}{4 \times 2}=\frac{3}{8} \mathrm{~s}$
Karnataka CET-2001
NUCLEAR PHYSICS
147745
The fraction of a sample of radioactive nuclei that remains undecayed in one mean life is
1 $\frac{1}{\mathrm{e}}$
2 $1-\frac{1}{\mathrm{e}}$
3 $\frac{1}{\mathrm{e}^{2}}$
4 $1-\frac{1}{\mathrm{e}^{2}}$
Explanation:
A According to equation of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Mean life time $(\mathrm{t})=\frac{1}{\lambda}$ Substituting value of mean life in equation (i), we get $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}$
J and K CET- 2010
NUCLEAR PHYSICS
147746
The activity of $1 \mathrm{mg}$ sample of ${ }_{37}^{90} \mathrm{Sr}$ whose half life is 28 years is (Given that Avogadro's number is $\mathbf{6 . 0 2} \times \mathbf{1 0}^{\mathbf{2 3}}$ )
147749
A radioactive substance decays to $(1 / 16)^{\text {th }}$ of its initial activity in 40 days. The half life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given, $\mathrm{t}=40 \text { days, } \frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{4}=\frac{40}{4}=10 \text { days }$
UP CPMT-2014
NUCLEAR PHYSICS
147750
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number of the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \alpha, 4 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D $\mathrm{Z}=90, \mathrm{Th}^{\mathrm{A}=228} \rightarrow \mathrm{Z}_{\mathrm{Z}^{\prime}=83} \mathrm{Bi}^{\mathrm{A}^{\prime}=212}$ Number of $\alpha$-particles emitted $\mathrm{n}_{\alpha}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{4}=\frac{228-212}{4}=4$ Number of $\beta$-particle emitted $\mathrm{n}_{\beta}=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ So, $4 \alpha$ and $1 \beta$ particles are emitted.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
147743
Three fourths of the active nuclei present in a radioactive sample decay in $3 / 4 \mathrm{~s}$. The half life of the sample is :
1 $\frac{3}{8} \mathrm{~s}$
2 $\frac{3}{4} \mathrm{~s}$
3 $\frac{1}{2} \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
A Given, Time $(\mathrm{t})=\frac{3}{4} \mathrm{~s}$ After radioactive sample decay remaining is $\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=2$ $\mathrm{T}_{1 / 2}=\frac{\mathrm{t}}{2}=\frac{3}{4 \times 2}=\frac{3}{8} \mathrm{~s}$
Karnataka CET-2001
NUCLEAR PHYSICS
147745
The fraction of a sample of radioactive nuclei that remains undecayed in one mean life is
1 $\frac{1}{\mathrm{e}}$
2 $1-\frac{1}{\mathrm{e}}$
3 $\frac{1}{\mathrm{e}^{2}}$
4 $1-\frac{1}{\mathrm{e}^{2}}$
Explanation:
A According to equation of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Mean life time $(\mathrm{t})=\frac{1}{\lambda}$ Substituting value of mean life in equation (i), we get $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}$
J and K CET- 2010
NUCLEAR PHYSICS
147746
The activity of $1 \mathrm{mg}$ sample of ${ }_{37}^{90} \mathrm{Sr}$ whose half life is 28 years is (Given that Avogadro's number is $\mathbf{6 . 0 2} \times \mathbf{1 0}^{\mathbf{2 3}}$ )
147749
A radioactive substance decays to $(1 / 16)^{\text {th }}$ of its initial activity in 40 days. The half life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given, $\mathrm{t}=40 \text { days, } \frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{4}=\frac{40}{4}=10 \text { days }$
UP CPMT-2014
NUCLEAR PHYSICS
147750
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number of the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \alpha, 4 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D $\mathrm{Z}=90, \mathrm{Th}^{\mathrm{A}=228} \rightarrow \mathrm{Z}_{\mathrm{Z}^{\prime}=83} \mathrm{Bi}^{\mathrm{A}^{\prime}=212}$ Number of $\alpha$-particles emitted $\mathrm{n}_{\alpha}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{4}=\frac{228-212}{4}=4$ Number of $\beta$-particle emitted $\mathrm{n}_{\beta}=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ So, $4 \alpha$ and $1 \beta$ particles are emitted.
147743
Three fourths of the active nuclei present in a radioactive sample decay in $3 / 4 \mathrm{~s}$. The half life of the sample is :
1 $\frac{3}{8} \mathrm{~s}$
2 $\frac{3}{4} \mathrm{~s}$
3 $\frac{1}{2} \mathrm{~s}$
4 $1 \mathrm{~s}$
Explanation:
A Given, Time $(\mathrm{t})=\frac{3}{4} \mathrm{~s}$ After radioactive sample decay remaining is $\frac{1}{4}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}$ $\frac{1}{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=2$ $\mathrm{T}_{1 / 2}=\frac{\mathrm{t}}{2}=\frac{3}{4 \times 2}=\frac{3}{8} \mathrm{~s}$
Karnataka CET-2001
NUCLEAR PHYSICS
147745
The fraction of a sample of radioactive nuclei that remains undecayed in one mean life is
1 $\frac{1}{\mathrm{e}}$
2 $1-\frac{1}{\mathrm{e}}$
3 $\frac{1}{\mathrm{e}^{2}}$
4 $1-\frac{1}{\mathrm{e}^{2}}$
Explanation:
A According to equation of radioactivity $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Mean life time $(\mathrm{t})=\frac{1}{\lambda}$ Substituting value of mean life in equation (i), we get $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \times \frac{1}{\lambda}}$ $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-1}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{\mathrm{e}}$
J and K CET- 2010
NUCLEAR PHYSICS
147746
The activity of $1 \mathrm{mg}$ sample of ${ }_{37}^{90} \mathrm{Sr}$ whose half life is 28 years is (Given that Avogadro's number is $\mathbf{6 . 0 2} \times \mathbf{1 0}^{\mathbf{2 3}}$ )
147749
A radioactive substance decays to $(1 / 16)^{\text {th }}$ of its initial activity in 40 days. The half life of the radioactive substance expressed in days is
1 2.5
2 5
3 10
4 20
Explanation:
C Given, $\mathrm{t}=40 \text { days, } \frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{1}{16}$ We know that, $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $4=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}$ $\mathrm{~T}_{1 / 2}=\frac{\mathrm{t}}{4}=\frac{40}{4}=10 \text { days }$
UP CPMT-2014
NUCLEAR PHYSICS
147750
When ${ }_{90} \mathrm{Th}^{228}$ transforms to ${ }_{83} \mathrm{Bi}^{212}$, then the number of the emitted $\alpha$ and $\beta$-particles is, respectively
1 $8 \alpha, 7 \beta$
2 $4 \alpha, 7 \beta$
3 $4 \alpha, 4 \beta$
4 $4 \alpha, 1 \beta$
Explanation:
D $\mathrm{Z}=90, \mathrm{Th}^{\mathrm{A}=228} \rightarrow \mathrm{Z}_{\mathrm{Z}^{\prime}=83} \mathrm{Bi}^{\mathrm{A}^{\prime}=212}$ Number of $\alpha$-particles emitted $\mathrm{n}_{\alpha}=\frac{\mathrm{A}-\mathrm{A}^{\prime}}{4}=\frac{228-212}{4}=4$ Number of $\beta$-particle emitted $\mathrm{n}_{\beta}=2 \mathrm{n}_{\alpha}-\mathrm{Z}+\mathrm{Z}^{\prime}$ $=2 \times 4-90+83$ $=1$ So, $4 \alpha$ and $1 \beta$ particles are emitted.