147734
The half life for $\alpha$ - decay of uranium ${ }_{92} \mathrm{U}^{228}$ is $4.47 \times 10^{8} \mathrm{yr}$. If a rock contains $60 \%$ of original ${ }_{92} \mathrm{U}^{228}$ atoms, then its age is [take $\log 6=0.778, \log 2=0.3$ ]
1 $1.2 \times 10^{7} \mathrm{yr}$
2 $3.3 \times 10^{8} \mathrm{yr}$
3 $4.2 \times 10^{9} \mathrm{yr}$
4 $6.5 \times 10^{9} \mathrm{yr}$
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=4.47 \times 10^{8} \mathrm{yr}$ According to the question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $2^{\mathrm{n}}=\frac{10}{6}$ Taking log both side, $\mathrm{n} \log 2=\log 10-\log 6$ $\mathrm{n} \times 0.3=1-0.778=0.22$ $\mathrm{n}=\frac{0.22}{0.3}=0.74$ $\text { So, } \quad \mathrm{t}=\mathrm{n} \mathrm{T}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ $=3.3 \times 10^{8} \mathrm{yr}$
VITEEE-2015
NUCLEAR PHYSICS
147735
The half life period of Radium is 3 minute. Its mean life time is
1 1.5 minute
2 $\frac{3}{0.6931}$ minute
3 6 minute
4 $(3 \times 0.6931)$ minute
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3$ minute As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}$ $\text { Mean life }(\tau)=\frac{1}{\lambda}$ $=\frac{3}{0.6931} \text { minute }$
VITEEE-2006
NUCLEAR PHYSICS
147728
In a radioactive reaction ${ }_{92} \mathrm{X}^{232} \rightarrow{ }_{82} \mathrm{X}^{204}$ the number of $\alpha$-particles emitted is :
1 7
2 6
3 5
4 4
Explanation:
A When an $\alpha$-particle is emitted from a nucleus, the resultant nucleus reduces its mass number by 4 units and its atomic number by 2 units. Loss in mass number $=232-204=28$ Therefore, number of $\alpha$-particle emitted $=\frac{28}{4}=7$
BCECE-2005
NUCLEAR PHYSICS
147740
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 According to radioactivity decay law $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days. $\frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substituting value of equation (i) in (ii), we get $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
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NUCLEAR PHYSICS
147734
The half life for $\alpha$ - decay of uranium ${ }_{92} \mathrm{U}^{228}$ is $4.47 \times 10^{8} \mathrm{yr}$. If a rock contains $60 \%$ of original ${ }_{92} \mathrm{U}^{228}$ atoms, then its age is [take $\log 6=0.778, \log 2=0.3$ ]
1 $1.2 \times 10^{7} \mathrm{yr}$
2 $3.3 \times 10^{8} \mathrm{yr}$
3 $4.2 \times 10^{9} \mathrm{yr}$
4 $6.5 \times 10^{9} \mathrm{yr}$
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=4.47 \times 10^{8} \mathrm{yr}$ According to the question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $2^{\mathrm{n}}=\frac{10}{6}$ Taking log both side, $\mathrm{n} \log 2=\log 10-\log 6$ $\mathrm{n} \times 0.3=1-0.778=0.22$ $\mathrm{n}=\frac{0.22}{0.3}=0.74$ $\text { So, } \quad \mathrm{t}=\mathrm{n} \mathrm{T}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ $=3.3 \times 10^{8} \mathrm{yr}$
VITEEE-2015
NUCLEAR PHYSICS
147735
The half life period of Radium is 3 minute. Its mean life time is
1 1.5 minute
2 $\frac{3}{0.6931}$ minute
3 6 minute
4 $(3 \times 0.6931)$ minute
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3$ minute As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}$ $\text { Mean life }(\tau)=\frac{1}{\lambda}$ $=\frac{3}{0.6931} \text { minute }$
VITEEE-2006
NUCLEAR PHYSICS
147728
In a radioactive reaction ${ }_{92} \mathrm{X}^{232} \rightarrow{ }_{82} \mathrm{X}^{204}$ the number of $\alpha$-particles emitted is :
1 7
2 6
3 5
4 4
Explanation:
A When an $\alpha$-particle is emitted from a nucleus, the resultant nucleus reduces its mass number by 4 units and its atomic number by 2 units. Loss in mass number $=232-204=28$ Therefore, number of $\alpha$-particle emitted $=\frac{28}{4}=7$
BCECE-2005
NUCLEAR PHYSICS
147740
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 According to radioactivity decay law $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days. $\frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substituting value of equation (i) in (ii), we get $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
147734
The half life for $\alpha$ - decay of uranium ${ }_{92} \mathrm{U}^{228}$ is $4.47 \times 10^{8} \mathrm{yr}$. If a rock contains $60 \%$ of original ${ }_{92} \mathrm{U}^{228}$ atoms, then its age is [take $\log 6=0.778, \log 2=0.3$ ]
1 $1.2 \times 10^{7} \mathrm{yr}$
2 $3.3 \times 10^{8} \mathrm{yr}$
3 $4.2 \times 10^{9} \mathrm{yr}$
4 $6.5 \times 10^{9} \mathrm{yr}$
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=4.47 \times 10^{8} \mathrm{yr}$ According to the question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $2^{\mathrm{n}}=\frac{10}{6}$ Taking log both side, $\mathrm{n} \log 2=\log 10-\log 6$ $\mathrm{n} \times 0.3=1-0.778=0.22$ $\mathrm{n}=\frac{0.22}{0.3}=0.74$ $\text { So, } \quad \mathrm{t}=\mathrm{n} \mathrm{T}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ $=3.3 \times 10^{8} \mathrm{yr}$
VITEEE-2015
NUCLEAR PHYSICS
147735
The half life period of Radium is 3 minute. Its mean life time is
1 1.5 minute
2 $\frac{3}{0.6931}$ minute
3 6 minute
4 $(3 \times 0.6931)$ minute
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3$ minute As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}$ $\text { Mean life }(\tau)=\frac{1}{\lambda}$ $=\frac{3}{0.6931} \text { minute }$
VITEEE-2006
NUCLEAR PHYSICS
147728
In a radioactive reaction ${ }_{92} \mathrm{X}^{232} \rightarrow{ }_{82} \mathrm{X}^{204}$ the number of $\alpha$-particles emitted is :
1 7
2 6
3 5
4 4
Explanation:
A When an $\alpha$-particle is emitted from a nucleus, the resultant nucleus reduces its mass number by 4 units and its atomic number by 2 units. Loss in mass number $=232-204=28$ Therefore, number of $\alpha$-particle emitted $=\frac{28}{4}=7$
BCECE-2005
NUCLEAR PHYSICS
147740
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 According to radioactivity decay law $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days. $\frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substituting value of equation (i) in (ii), we get $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
147734
The half life for $\alpha$ - decay of uranium ${ }_{92} \mathrm{U}^{228}$ is $4.47 \times 10^{8} \mathrm{yr}$. If a rock contains $60 \%$ of original ${ }_{92} \mathrm{U}^{228}$ atoms, then its age is [take $\log 6=0.778, \log 2=0.3$ ]
1 $1.2 \times 10^{7} \mathrm{yr}$
2 $3.3 \times 10^{8} \mathrm{yr}$
3 $4.2 \times 10^{9} \mathrm{yr}$
4 $6.5 \times 10^{9} \mathrm{yr}$
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=4.47 \times 10^{8} \mathrm{yr}$ According to the question, $\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}=\frac{60}{100}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $2^{\mathrm{n}}=\frac{10}{6}$ Taking log both side, $\mathrm{n} \log 2=\log 10-\log 6$ $\mathrm{n} \times 0.3=1-0.778=0.22$ $\mathrm{n}=\frac{0.22}{0.3}=0.74$ $\text { So, } \quad \mathrm{t}=\mathrm{n} \mathrm{T}_{1 / 2}=0.74 \times 4.47 \times 10^{8}$ $=3.3 \times 10^{8} \mathrm{yr}$
VITEEE-2015
NUCLEAR PHYSICS
147735
The half life period of Radium is 3 minute. Its mean life time is
1 1.5 minute
2 $\frac{3}{0.6931}$ minute
3 6 minute
4 $(3 \times 0.6931)$ minute
Explanation:
B Given, half life $\left(\mathrm{T}_{1 / 2}\right)=3$ minute As we know that, $\mathrm{T}_{1 / 2}=\frac{\ln 2}{\lambda}$ $\lambda=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}$ $\text { Mean life }(\tau)=\frac{1}{\lambda}$ $=\frac{3}{0.6931} \text { minute }$
VITEEE-2006
NUCLEAR PHYSICS
147728
In a radioactive reaction ${ }_{92} \mathrm{X}^{232} \rightarrow{ }_{82} \mathrm{X}^{204}$ the number of $\alpha$-particles emitted is :
1 7
2 6
3 5
4 4
Explanation:
A When an $\alpha$-particle is emitted from a nucleus, the resultant nucleus reduces its mass number by 4 units and its atomic number by 2 units. Loss in mass number $=232-204=28$ Therefore, number of $\alpha$-particle emitted $=\frac{28}{4}=7$
BCECE-2005
NUCLEAR PHYSICS
147740
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 According to radioactivity decay law $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ Activity of a radioactive sample decreases to one third of its original value in 3 days. $\frac{\mathrm{N}_{0}}{3}=\mathrm{N}_{0} \mathrm{e}^{-3 \lambda}$ $\frac{1}{3}=\mathrm{e}^{-3 \lambda}$ Let activity in 9 days be $\mathrm{N}^{\prime}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda \times 9}$ $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\mathrm{e}^{-3 \lambda}\right)^{3}$ Substituting value of equation (i) in (ii), we get $\frac{\mathrm{N}^{\prime}}{\mathrm{N}_{0}}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$
