147688
Radium has a half-life of $5 \mathrm{yr}$. The probability of decay of a radium nucleus in $10 \mathrm{yr}$ is
1 $50 \%$
2 $75 \%$
3 $100 \%$
4 $60 \%$
5 $25 \%$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{yr}$. Time $(\mathrm{t})=10 \mathrm{yr}$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}=\left(\frac{1}{2}\right)^{10 / 5}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=25 \%$ Probability of remaining portion of radium nucleus $=$ $(100-25) \%=75 \%$
Kerala CEE 2007
NUCLEAR PHYSICS
147689
The half-life of radon is 3.8 days. How many radon will be left out of $1024 \mathrm{mg}$ after 38 days:
1 $1 \mathrm{mg}$
2 $2 \mathrm{mg}$
3 $3 \mathrm{mg}$
4 $4 \mathrm{mg}$
5 $7 \mathrm{mg}$
Explanation:
A Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ days Time $(\mathrm{t})=38$ days Mass of radon $\left(\mathrm{m}_{\mathrm{o}}\right)=1024 \mathrm{mg}$ Remaining mass of radon $(\mathrm{m})=$ ? As we know, Half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\lambda=\frac{0.693}{3.8}=0.182$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{m}=1024 \times \mathrm{e}^{-(0.182) \times 38}$ $\mathrm{m}=1024 \times \mathrm{e}^{-6.916}$ $\mathrm{m}=1024 \times 0.000991$ $\mathrm{m}=1 \mathrm{mg}$
Kerala CEE 2006
NUCLEAR PHYSICS
147690
The number of $\alpha$-particles and $\beta$-particles respectively emitted in the reaction ${ }_{88} \mathbf{A}^{196} \rightarrow{ }_{78} \mathbf{B}^{164}$ are :
1 8 and 8
2 8 and 6
3 6 and 8
4 6 and 6
5 8 and 4
Explanation:
B Let, the number of $\alpha$-particle is $x$ and number of $\beta$-particle is $y$. Given reaction can be written as - ${ }_{88} \mathrm{~A}^{196} \rightarrow{ }_{78} \mathrm{~B}^{164}+\mathrm{x}\left({ }_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ According to conservation of atomic number, we get - $88=78+2 x-y$ $2 x-y=10$ According to conservation mass number, we get - $196=164+4 x+0$ $4 x=32$ $x=8$ Putting value of $\mathrm{x}$ in equation (i), we get - $2 \times 8-y=10$ $y=6$ Hence, $8 \alpha$-particles and $6 \beta$-particles are emitted.
Kerala CEE 2005
NUCLEAR PHYSICS
147691
$C^{14}$ has half-life 5700 year. At the end of 11400 years. The actual amount left is:
1 0.5 of original amount
2 0.25 of original amount
3 0.125 of original amount
4 0.0625 of original amount
5 0.03125 of original amount
Explanation:
B Given, Half life $\left(T_{1 / 2}\right)=5700$ years Total time $(t)=11400$ years $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{11400}{5700}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}=0.25 \mathrm{~N}_{0}$ Hence, the actual amount left is 0.25 of original amount.
147688
Radium has a half-life of $5 \mathrm{yr}$. The probability of decay of a radium nucleus in $10 \mathrm{yr}$ is
1 $50 \%$
2 $75 \%$
3 $100 \%$
4 $60 \%$
5 $25 \%$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{yr}$. Time $(\mathrm{t})=10 \mathrm{yr}$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}=\left(\frac{1}{2}\right)^{10 / 5}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=25 \%$ Probability of remaining portion of radium nucleus $=$ $(100-25) \%=75 \%$
Kerala CEE 2007
NUCLEAR PHYSICS
147689
The half-life of radon is 3.8 days. How many radon will be left out of $1024 \mathrm{mg}$ after 38 days:
1 $1 \mathrm{mg}$
2 $2 \mathrm{mg}$
3 $3 \mathrm{mg}$
4 $4 \mathrm{mg}$
5 $7 \mathrm{mg}$
Explanation:
A Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ days Time $(\mathrm{t})=38$ days Mass of radon $\left(\mathrm{m}_{\mathrm{o}}\right)=1024 \mathrm{mg}$ Remaining mass of radon $(\mathrm{m})=$ ? As we know, Half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\lambda=\frac{0.693}{3.8}=0.182$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{m}=1024 \times \mathrm{e}^{-(0.182) \times 38}$ $\mathrm{m}=1024 \times \mathrm{e}^{-6.916}$ $\mathrm{m}=1024 \times 0.000991$ $\mathrm{m}=1 \mathrm{mg}$
Kerala CEE 2006
NUCLEAR PHYSICS
147690
The number of $\alpha$-particles and $\beta$-particles respectively emitted in the reaction ${ }_{88} \mathbf{A}^{196} \rightarrow{ }_{78} \mathbf{B}^{164}$ are :
1 8 and 8
2 8 and 6
3 6 and 8
4 6 and 6
5 8 and 4
Explanation:
B Let, the number of $\alpha$-particle is $x$ and number of $\beta$-particle is $y$. Given reaction can be written as - ${ }_{88} \mathrm{~A}^{196} \rightarrow{ }_{78} \mathrm{~B}^{164}+\mathrm{x}\left({ }_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ According to conservation of atomic number, we get - $88=78+2 x-y$ $2 x-y=10$ According to conservation mass number, we get - $196=164+4 x+0$ $4 x=32$ $x=8$ Putting value of $\mathrm{x}$ in equation (i), we get - $2 \times 8-y=10$ $y=6$ Hence, $8 \alpha$-particles and $6 \beta$-particles are emitted.
Kerala CEE 2005
NUCLEAR PHYSICS
147691
$C^{14}$ has half-life 5700 year. At the end of 11400 years. The actual amount left is:
1 0.5 of original amount
2 0.25 of original amount
3 0.125 of original amount
4 0.0625 of original amount
5 0.03125 of original amount
Explanation:
B Given, Half life $\left(T_{1 / 2}\right)=5700$ years Total time $(t)=11400$ years $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{11400}{5700}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}=0.25 \mathrm{~N}_{0}$ Hence, the actual amount left is 0.25 of original amount.
147688
Radium has a half-life of $5 \mathrm{yr}$. The probability of decay of a radium nucleus in $10 \mathrm{yr}$ is
1 $50 \%$
2 $75 \%$
3 $100 \%$
4 $60 \%$
5 $25 \%$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{yr}$. Time $(\mathrm{t})=10 \mathrm{yr}$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}=\left(\frac{1}{2}\right)^{10 / 5}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=25 \%$ Probability of remaining portion of radium nucleus $=$ $(100-25) \%=75 \%$
Kerala CEE 2007
NUCLEAR PHYSICS
147689
The half-life of radon is 3.8 days. How many radon will be left out of $1024 \mathrm{mg}$ after 38 days:
1 $1 \mathrm{mg}$
2 $2 \mathrm{mg}$
3 $3 \mathrm{mg}$
4 $4 \mathrm{mg}$
5 $7 \mathrm{mg}$
Explanation:
A Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ days Time $(\mathrm{t})=38$ days Mass of radon $\left(\mathrm{m}_{\mathrm{o}}\right)=1024 \mathrm{mg}$ Remaining mass of radon $(\mathrm{m})=$ ? As we know, Half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\lambda=\frac{0.693}{3.8}=0.182$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{m}=1024 \times \mathrm{e}^{-(0.182) \times 38}$ $\mathrm{m}=1024 \times \mathrm{e}^{-6.916}$ $\mathrm{m}=1024 \times 0.000991$ $\mathrm{m}=1 \mathrm{mg}$
Kerala CEE 2006
NUCLEAR PHYSICS
147690
The number of $\alpha$-particles and $\beta$-particles respectively emitted in the reaction ${ }_{88} \mathbf{A}^{196} \rightarrow{ }_{78} \mathbf{B}^{164}$ are :
1 8 and 8
2 8 and 6
3 6 and 8
4 6 and 6
5 8 and 4
Explanation:
B Let, the number of $\alpha$-particle is $x$ and number of $\beta$-particle is $y$. Given reaction can be written as - ${ }_{88} \mathrm{~A}^{196} \rightarrow{ }_{78} \mathrm{~B}^{164}+\mathrm{x}\left({ }_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ According to conservation of atomic number, we get - $88=78+2 x-y$ $2 x-y=10$ According to conservation mass number, we get - $196=164+4 x+0$ $4 x=32$ $x=8$ Putting value of $\mathrm{x}$ in equation (i), we get - $2 \times 8-y=10$ $y=6$ Hence, $8 \alpha$-particles and $6 \beta$-particles are emitted.
Kerala CEE 2005
NUCLEAR PHYSICS
147691
$C^{14}$ has half-life 5700 year. At the end of 11400 years. The actual amount left is:
1 0.5 of original amount
2 0.25 of original amount
3 0.125 of original amount
4 0.0625 of original amount
5 0.03125 of original amount
Explanation:
B Given, Half life $\left(T_{1 / 2}\right)=5700$ years Total time $(t)=11400$ years $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{11400}{5700}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}=0.25 \mathrm{~N}_{0}$ Hence, the actual amount left is 0.25 of original amount.
147688
Radium has a half-life of $5 \mathrm{yr}$. The probability of decay of a radium nucleus in $10 \mathrm{yr}$ is
1 $50 \%$
2 $75 \%$
3 $100 \%$
4 $60 \%$
5 $25 \%$
Explanation:
B Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{yr}$. Time $(\mathrm{t})=10 \mathrm{yr}$. $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}_{1 / 2}}=\left(\frac{1}{2}\right)^{10 / 5}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}=25 \%$ Probability of remaining portion of radium nucleus $=$ $(100-25) \%=75 \%$
Kerala CEE 2007
NUCLEAR PHYSICS
147689
The half-life of radon is 3.8 days. How many radon will be left out of $1024 \mathrm{mg}$ after 38 days:
1 $1 \mathrm{mg}$
2 $2 \mathrm{mg}$
3 $3 \mathrm{mg}$
4 $4 \mathrm{mg}$
5 $7 \mathrm{mg}$
Explanation:
A Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3.8$ days Time $(\mathrm{t})=38$ days Mass of radon $\left(\mathrm{m}_{\mathrm{o}}\right)=1024 \mathrm{mg}$ Remaining mass of radon $(\mathrm{m})=$ ? As we know, Half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\lambda=\frac{0.693}{3.8}=0.182$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{\mathrm{o}} \mathrm{e}^{-\lambda \mathrm{t}}$ $\mathrm{m}=1024 \times \mathrm{e}^{-(0.182) \times 38}$ $\mathrm{m}=1024 \times \mathrm{e}^{-6.916}$ $\mathrm{m}=1024 \times 0.000991$ $\mathrm{m}=1 \mathrm{mg}$
Kerala CEE 2006
NUCLEAR PHYSICS
147690
The number of $\alpha$-particles and $\beta$-particles respectively emitted in the reaction ${ }_{88} \mathbf{A}^{196} \rightarrow{ }_{78} \mathbf{B}^{164}$ are :
1 8 and 8
2 8 and 6
3 6 and 8
4 6 and 6
5 8 and 4
Explanation:
B Let, the number of $\alpha$-particle is $x$ and number of $\beta$-particle is $y$. Given reaction can be written as - ${ }_{88} \mathrm{~A}^{196} \rightarrow{ }_{78} \mathrm{~B}^{164}+\mathrm{x}\left({ }_{2} \mathrm{He}^{4}\right)+\mathrm{y}\left({ }_{-1} \mathrm{e}^{0}\right)$ According to conservation of atomic number, we get - $88=78+2 x-y$ $2 x-y=10$ According to conservation mass number, we get - $196=164+4 x+0$ $4 x=32$ $x=8$ Putting value of $\mathrm{x}$ in equation (i), we get - $2 \times 8-y=10$ $y=6$ Hence, $8 \alpha$-particles and $6 \beta$-particles are emitted.
Kerala CEE 2005
NUCLEAR PHYSICS
147691
$C^{14}$ has half-life 5700 year. At the end of 11400 years. The actual amount left is:
1 0.5 of original amount
2 0.25 of original amount
3 0.125 of original amount
4 0.0625 of original amount
5 0.03125 of original amount
Explanation:
B Given, Half life $\left(T_{1 / 2}\right)=5700$ years Total time $(t)=11400$ years $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\mathrm{N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{\frac{11400}{5700}}$ $\mathrm{~N}=\mathrm{N}_{0} \times\left(\frac{1}{2}\right)^{2}$ $\mathrm{~N}=0.25 \mathrm{~N}_{0}$ Hence, the actual amount left is 0.25 of original amount.
