NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147629
Half life of radioactive element is $5 \mathrm{~min}$. At the end of $20 \mathrm{~min}$ its disintegrated.
1 6.25
2 75
3 25
4 93.75 $\%$ quality is
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ So, Total time $(\mathrm{t})=20 \mathrm{~min}$ Number of half-life $(n)=\frac{\text { Total time }(t)}{\operatorname{Half} \operatorname{life}\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{20}{5}=4$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Disintegrated $=\left(\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{1}{16}\right) \times 100$ $=\frac{15}{16} \times 100$ $=93.75 \%$
GUJCET 2017
NUCLEAR PHYSICS
147630
Cobalt-57 is radioactive, emitting $\beta$-particles. The half life for this is 270 days. If $100 \mathrm{mg}$ of this is kept in an open container, then the mass of Cobalt-57 after 540 days will be
1 $50 \mathrm{mg}$
2 $\left(\frac{50}{\sqrt{2}}\right) \mathrm{mg}$
3 $25 \mathrm{mg}$
4 zero
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=270$ days Initial amount of cobalt-57 $(\mathrm{N})=100 \mathrm{mg}$ Time of observation $(t)=540$ days So, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{540}{270}=2$ Thus, the amount of cobalt-57 remaining $(\mathrm{N})=\frac{\mathrm{N}_{0}}{2^{\mathrm{n}}}$ $=\frac{100}{2^{2}}=25 \mathrm{mg}$
JCECE-2017
NUCLEAR PHYSICS
147631
The half life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially they have the same number of atoms. Then
1 $\mathrm{X}$ and $\mathrm{Y}$ decay at same rate always
2 $X$ will decay faster than $Y$
3 Y will decay faster than $\mathrm{X}$
4 $\mathrm{X}$ and $\mathrm{Y}$ have same decay rate initially
Explanation:
C Half life of $\mathrm{X}\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=$ Mean life of $\mathrm{Y}(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ Since, $\quad \ln 2=0.693 \lt 1$ So, $\quad \lambda_{\mathrm{Y}}>\lambda_{\mathrm{X}}$ Higher is the decay constant higher is the rate of decay. So, Y will decay faster than $\mathrm{X}$.
JCECE-2017
NUCLEAR PHYSICS
147632
$8 \mathrm{~g}$ of $\mathrm{Cu}^{66}$ undergoes radioactive decay and after 15 minutes only $1 \mathrm{~g}$ remains. The half-life, in minutes, is then
1 $15 \ln (2) / \ln (8)$
2 $15 \ln (8) / \ln (2)$
3 $15 / 8$
4 $8 / 15$
5 15 in (2)
Explanation:
A Given, Initial mass of substance $\left(\mathrm{m}_{0}\right)=8 \mathrm{~g}$ Remaining mass of substance $(\mathrm{m})=1 \mathrm{~g}$ Time of decay $(t)=15 \mathrm{~min}$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $1=8 \mathrm{e}^{-\lambda \times 15}$ $\mathrm{e}^{\lambda \times 15}=8$ Taking $\log$ on both side - $15 \lambda=\ln 8$ $\lambda=\frac{\ln 8}{15}$ So, half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $=\frac{15 \ln 2}{\ln 8}$
147629
Half life of radioactive element is $5 \mathrm{~min}$. At the end of $20 \mathrm{~min}$ its disintegrated.
1 6.25
2 75
3 25
4 93.75 $\%$ quality is
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ So, Total time $(\mathrm{t})=20 \mathrm{~min}$ Number of half-life $(n)=\frac{\text { Total time }(t)}{\operatorname{Half} \operatorname{life}\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{20}{5}=4$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Disintegrated $=\left(\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{1}{16}\right) \times 100$ $=\frac{15}{16} \times 100$ $=93.75 \%$
GUJCET 2017
NUCLEAR PHYSICS
147630
Cobalt-57 is radioactive, emitting $\beta$-particles. The half life for this is 270 days. If $100 \mathrm{mg}$ of this is kept in an open container, then the mass of Cobalt-57 after 540 days will be
1 $50 \mathrm{mg}$
2 $\left(\frac{50}{\sqrt{2}}\right) \mathrm{mg}$
3 $25 \mathrm{mg}$
4 zero
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=270$ days Initial amount of cobalt-57 $(\mathrm{N})=100 \mathrm{mg}$ Time of observation $(t)=540$ days So, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{540}{270}=2$ Thus, the amount of cobalt-57 remaining $(\mathrm{N})=\frac{\mathrm{N}_{0}}{2^{\mathrm{n}}}$ $=\frac{100}{2^{2}}=25 \mathrm{mg}$
JCECE-2017
NUCLEAR PHYSICS
147631
The half life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially they have the same number of atoms. Then
1 $\mathrm{X}$ and $\mathrm{Y}$ decay at same rate always
2 $X$ will decay faster than $Y$
3 Y will decay faster than $\mathrm{X}$
4 $\mathrm{X}$ and $\mathrm{Y}$ have same decay rate initially
Explanation:
C Half life of $\mathrm{X}\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=$ Mean life of $\mathrm{Y}(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ Since, $\quad \ln 2=0.693 \lt 1$ So, $\quad \lambda_{\mathrm{Y}}>\lambda_{\mathrm{X}}$ Higher is the decay constant higher is the rate of decay. So, Y will decay faster than $\mathrm{X}$.
JCECE-2017
NUCLEAR PHYSICS
147632
$8 \mathrm{~g}$ of $\mathrm{Cu}^{66}$ undergoes radioactive decay and after 15 minutes only $1 \mathrm{~g}$ remains. The half-life, in minutes, is then
1 $15 \ln (2) / \ln (8)$
2 $15 \ln (8) / \ln (2)$
3 $15 / 8$
4 $8 / 15$
5 15 in (2)
Explanation:
A Given, Initial mass of substance $\left(\mathrm{m}_{0}\right)=8 \mathrm{~g}$ Remaining mass of substance $(\mathrm{m})=1 \mathrm{~g}$ Time of decay $(t)=15 \mathrm{~min}$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $1=8 \mathrm{e}^{-\lambda \times 15}$ $\mathrm{e}^{\lambda \times 15}=8$ Taking $\log$ on both side - $15 \lambda=\ln 8$ $\lambda=\frac{\ln 8}{15}$ So, half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $=\frac{15 \ln 2}{\ln 8}$
147629
Half life of radioactive element is $5 \mathrm{~min}$. At the end of $20 \mathrm{~min}$ its disintegrated.
1 6.25
2 75
3 25
4 93.75 $\%$ quality is
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ So, Total time $(\mathrm{t})=20 \mathrm{~min}$ Number of half-life $(n)=\frac{\text { Total time }(t)}{\operatorname{Half} \operatorname{life}\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{20}{5}=4$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Disintegrated $=\left(\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{1}{16}\right) \times 100$ $=\frac{15}{16} \times 100$ $=93.75 \%$
GUJCET 2017
NUCLEAR PHYSICS
147630
Cobalt-57 is radioactive, emitting $\beta$-particles. The half life for this is 270 days. If $100 \mathrm{mg}$ of this is kept in an open container, then the mass of Cobalt-57 after 540 days will be
1 $50 \mathrm{mg}$
2 $\left(\frac{50}{\sqrt{2}}\right) \mathrm{mg}$
3 $25 \mathrm{mg}$
4 zero
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=270$ days Initial amount of cobalt-57 $(\mathrm{N})=100 \mathrm{mg}$ Time of observation $(t)=540$ days So, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{540}{270}=2$ Thus, the amount of cobalt-57 remaining $(\mathrm{N})=\frac{\mathrm{N}_{0}}{2^{\mathrm{n}}}$ $=\frac{100}{2^{2}}=25 \mathrm{mg}$
JCECE-2017
NUCLEAR PHYSICS
147631
The half life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially they have the same number of atoms. Then
1 $\mathrm{X}$ and $\mathrm{Y}$ decay at same rate always
2 $X$ will decay faster than $Y$
3 Y will decay faster than $\mathrm{X}$
4 $\mathrm{X}$ and $\mathrm{Y}$ have same decay rate initially
Explanation:
C Half life of $\mathrm{X}\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=$ Mean life of $\mathrm{Y}(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ Since, $\quad \ln 2=0.693 \lt 1$ So, $\quad \lambda_{\mathrm{Y}}>\lambda_{\mathrm{X}}$ Higher is the decay constant higher is the rate of decay. So, Y will decay faster than $\mathrm{X}$.
JCECE-2017
NUCLEAR PHYSICS
147632
$8 \mathrm{~g}$ of $\mathrm{Cu}^{66}$ undergoes radioactive decay and after 15 minutes only $1 \mathrm{~g}$ remains. The half-life, in minutes, is then
1 $15 \ln (2) / \ln (8)$
2 $15 \ln (8) / \ln (2)$
3 $15 / 8$
4 $8 / 15$
5 15 in (2)
Explanation:
A Given, Initial mass of substance $\left(\mathrm{m}_{0}\right)=8 \mathrm{~g}$ Remaining mass of substance $(\mathrm{m})=1 \mathrm{~g}$ Time of decay $(t)=15 \mathrm{~min}$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $1=8 \mathrm{e}^{-\lambda \times 15}$ $\mathrm{e}^{\lambda \times 15}=8$ Taking $\log$ on both side - $15 \lambda=\ln 8$ $\lambda=\frac{\ln 8}{15}$ So, half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $=\frac{15 \ln 2}{\ln 8}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
147629
Half life of radioactive element is $5 \mathrm{~min}$. At the end of $20 \mathrm{~min}$ its disintegrated.
1 6.25
2 75
3 25
4 93.75 $\%$ quality is
Explanation:
D Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=5 \mathrm{~min}$ So, Total time $(\mathrm{t})=20 \mathrm{~min}$ Number of half-life $(n)=\frac{\text { Total time }(t)}{\operatorname{Half} \operatorname{life}\left(T_{1 / 2}\right)}$ $\mathrm{n}=\frac{20}{5}=4$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{n}}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$ $\mathrm{N}=\frac{\mathrm{N}_{0}}{16}$ Disintegrated $=\left(\frac{\mathrm{N}_{0}-\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{\mathrm{N}}{\mathrm{N}_{0}}\right) \times 100$ $=\left(1-\frac{1}{16}\right) \times 100$ $=\frac{15}{16} \times 100$ $=93.75 \%$
GUJCET 2017
NUCLEAR PHYSICS
147630
Cobalt-57 is radioactive, emitting $\beta$-particles. The half life for this is 270 days. If $100 \mathrm{mg}$ of this is kept in an open container, then the mass of Cobalt-57 after 540 days will be
1 $50 \mathrm{mg}$
2 $\left(\frac{50}{\sqrt{2}}\right) \mathrm{mg}$
3 $25 \mathrm{mg}$
4 zero
Explanation:
C Given, Half life $\left(\mathrm{T}_{1 / 2}\right)=270$ days Initial amount of cobalt-57 $(\mathrm{N})=100 \mathrm{mg}$ Time of observation $(t)=540$ days So, $\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}=\frac{540}{270}=2$ Thus, the amount of cobalt-57 remaining $(\mathrm{N})=\frac{\mathrm{N}_{0}}{2^{\mathrm{n}}}$ $=\frac{100}{2^{2}}=25 \mathrm{mg}$
JCECE-2017
NUCLEAR PHYSICS
147631
The half life period of a radioactive element $X$ is same as the mean life time of another radioactive element $Y$. Initially they have the same number of atoms. Then
1 $\mathrm{X}$ and $\mathrm{Y}$ decay at same rate always
2 $X$ will decay faster than $Y$
3 Y will decay faster than $\mathrm{X}$
4 $\mathrm{X}$ and $\mathrm{Y}$ have same decay rate initially
Explanation:
C Half life of $\mathrm{X}\left(\mathrm{T}_{1 / 2}\right)_{\mathrm{X}}=$ Mean life of $\mathrm{Y}(\tau)_{\mathrm{Y}}$ $\frac{\ln 2}{\lambda_{\mathrm{X}}}=\frac{1}{\lambda_{\mathrm{Y}}}$ Since, $\quad \ln 2=0.693 \lt 1$ So, $\quad \lambda_{\mathrm{Y}}>\lambda_{\mathrm{X}}$ Higher is the decay constant higher is the rate of decay. So, Y will decay faster than $\mathrm{X}$.
JCECE-2017
NUCLEAR PHYSICS
147632
$8 \mathrm{~g}$ of $\mathrm{Cu}^{66}$ undergoes radioactive decay and after 15 minutes only $1 \mathrm{~g}$ remains. The half-life, in minutes, is then
1 $15 \ln (2) / \ln (8)$
2 $15 \ln (8) / \ln (2)$
3 $15 / 8$
4 $8 / 15$
5 15 in (2)
Explanation:
A Given, Initial mass of substance $\left(\mathrm{m}_{0}\right)=8 \mathrm{~g}$ Remaining mass of substance $(\mathrm{m})=1 \mathrm{~g}$ Time of decay $(t)=15 \mathrm{~min}$ According to radioactive decay law - $\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ $1=8 \mathrm{e}^{-\lambda \times 15}$ $\mathrm{e}^{\lambda \times 15}=8$ Taking $\log$ on both side - $15 \lambda=\ln 8$ $\lambda=\frac{\ln 8}{15}$ So, half life $\left(\mathrm{T}_{1 / 2}\right)=\frac{\ln 2}{\lambda}$ $=\frac{15 \ln 2}{\ln 8}$