NEET Test Series from KOTA - 10 Papers In MS WORD
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NUCLEAR PHYSICS
147711
Assertion: Cobalt-60 is useful in cancer therapy. Reason: Cobalt-60 is a source of $\gamma$-radiations capable of killing cancerous cells.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A Cobalt-60 is a source of $\gamma$-radiation, which are used in curing cancer by destroying the cancerous cell. So, cobalt-60 is useful in cancer therapy.
AIIMS-2006
NUCLEAR PHYSICS
147718
In nucleus of mass number $A$, originally at rest, emits an $\alpha$-particle with speed $v$. The daughter nucleus recoils with a speed :
1 $\frac{2 \mathrm{v}}{\mathrm{A}+4}$
2 $\frac{4 \mathrm{v}}{\mathrm{A}+4}$
3 $\frac{4 \mathrm{v}}{\mathrm{A}-4}$
4 $\frac{2 \mathrm{v}}{\mathrm{A}-4}$
Explanation:
C As, $\mathrm{f}_{\text {net }}=0$, therefore, momentum is conserved. $\overrightarrow{\mathrm{P}}_{1}+\overrightarrow{\mathrm{P}}_{2}=0$ $\left|\overrightarrow{\mathrm{P}}_{1}\right|=\left|\overrightarrow{\mathrm{P}}_{2}\right|$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\mathrm{~m}(\mathrm{~A}-4) \mathrm{v}_{1}=4 \mathrm{mv}$ $\mathrm{v}_{1}=\frac{4 \mathrm{v}}{\mathrm{A}-4}$
AIIMS-2004
NUCLEAR PHYSICS
147720
A radioactive material decays by simultaneous emission of two particles with half-lives $1620 \mathrm{yr}$ and $810 \mathrm{yr}$ respectively. The time in years after which one-fourth of material remains, is-
1 $1080 \mathrm{yr}$
2 $2340 \mathrm{yr}$
3 $4860 \mathrm{yr}$
4 $3240 \mathrm{yr}$
Explanation:
A Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ The number of half lives (n) $=\frac{\text { timeof decay }}{\text { effectivehalf }- \text { life }}$ Relation between effective disintegration constant $(\lambda)$ and half life $\left(\mathrm{T}_{1 / 2}\right)$. $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\therefore \quad\lambda_{1}+\lambda_{2}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{1}}+\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{2}}$ Effective half life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}=\frac{1}{1620}+\frac{1}{810}$ $\frac{1}{\mathrm{~T}}=\frac{1+2}{1620}$ $\mathrm{~T}=540 \mathrm{yr}$ $\therefore \quad \mathrm{n}=\frac{\mathrm{T}}{540}$ $\therefore \quad \mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{t}}{540}=2$ $\text { So, } \quad \mathrm{t}=2 \times 540=1080 \mathrm{yr}$
BCECE-2015
NUCLEAR PHYSICS
147724
What is the disintegration constant of radon, if the number of its atoms diminishes by $18 \%$ in $24 \mathrm{~h}$ ?
1 $2.1 \times 10^{-3} \mathrm{~s}^{-1}$
2 $2.1 \times 10^{-5} \mathrm{~s}^{-1}$
3 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
4 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
Explanation:
D Undisintegrated part $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=(100-18) \%$ $=82 \%$ According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{82}{100}=\mathrm{e}^{-(24 \times 60 \times 60 \times \lambda)}$ $24 \times 60 \times 60 \times \lambda=\ln \left(\frac{100}{82}\right)$ $\lambda=2.2 \times 10^{-6} \mathrm{~s}^{-1}$
147711
Assertion: Cobalt-60 is useful in cancer therapy. Reason: Cobalt-60 is a source of $\gamma$-radiations capable of killing cancerous cells.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A Cobalt-60 is a source of $\gamma$-radiation, which are used in curing cancer by destroying the cancerous cell. So, cobalt-60 is useful in cancer therapy.
AIIMS-2006
NUCLEAR PHYSICS
147718
In nucleus of mass number $A$, originally at rest, emits an $\alpha$-particle with speed $v$. The daughter nucleus recoils with a speed :
1 $\frac{2 \mathrm{v}}{\mathrm{A}+4}$
2 $\frac{4 \mathrm{v}}{\mathrm{A}+4}$
3 $\frac{4 \mathrm{v}}{\mathrm{A}-4}$
4 $\frac{2 \mathrm{v}}{\mathrm{A}-4}$
Explanation:
C As, $\mathrm{f}_{\text {net }}=0$, therefore, momentum is conserved. $\overrightarrow{\mathrm{P}}_{1}+\overrightarrow{\mathrm{P}}_{2}=0$ $\left|\overrightarrow{\mathrm{P}}_{1}\right|=\left|\overrightarrow{\mathrm{P}}_{2}\right|$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\mathrm{~m}(\mathrm{~A}-4) \mathrm{v}_{1}=4 \mathrm{mv}$ $\mathrm{v}_{1}=\frac{4 \mathrm{v}}{\mathrm{A}-4}$
AIIMS-2004
NUCLEAR PHYSICS
147720
A radioactive material decays by simultaneous emission of two particles with half-lives $1620 \mathrm{yr}$ and $810 \mathrm{yr}$ respectively. The time in years after which one-fourth of material remains, is-
1 $1080 \mathrm{yr}$
2 $2340 \mathrm{yr}$
3 $4860 \mathrm{yr}$
4 $3240 \mathrm{yr}$
Explanation:
A Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ The number of half lives (n) $=\frac{\text { timeof decay }}{\text { effectivehalf }- \text { life }}$ Relation between effective disintegration constant $(\lambda)$ and half life $\left(\mathrm{T}_{1 / 2}\right)$. $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\therefore \quad\lambda_{1}+\lambda_{2}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{1}}+\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{2}}$ Effective half life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}=\frac{1}{1620}+\frac{1}{810}$ $\frac{1}{\mathrm{~T}}=\frac{1+2}{1620}$ $\mathrm{~T}=540 \mathrm{yr}$ $\therefore \quad \mathrm{n}=\frac{\mathrm{T}}{540}$ $\therefore \quad \mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{t}}{540}=2$ $\text { So, } \quad \mathrm{t}=2 \times 540=1080 \mathrm{yr}$
BCECE-2015
NUCLEAR PHYSICS
147724
What is the disintegration constant of radon, if the number of its atoms diminishes by $18 \%$ in $24 \mathrm{~h}$ ?
1 $2.1 \times 10^{-3} \mathrm{~s}^{-1}$
2 $2.1 \times 10^{-5} \mathrm{~s}^{-1}$
3 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
4 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
Explanation:
D Undisintegrated part $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=(100-18) \%$ $=82 \%$ According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{82}{100}=\mathrm{e}^{-(24 \times 60 \times 60 \times \lambda)}$ $24 \times 60 \times 60 \times \lambda=\ln \left(\frac{100}{82}\right)$ $\lambda=2.2 \times 10^{-6} \mathrm{~s}^{-1}$
147711
Assertion: Cobalt-60 is useful in cancer therapy. Reason: Cobalt-60 is a source of $\gamma$-radiations capable of killing cancerous cells.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A Cobalt-60 is a source of $\gamma$-radiation, which are used in curing cancer by destroying the cancerous cell. So, cobalt-60 is useful in cancer therapy.
AIIMS-2006
NUCLEAR PHYSICS
147718
In nucleus of mass number $A$, originally at rest, emits an $\alpha$-particle with speed $v$. The daughter nucleus recoils with a speed :
1 $\frac{2 \mathrm{v}}{\mathrm{A}+4}$
2 $\frac{4 \mathrm{v}}{\mathrm{A}+4}$
3 $\frac{4 \mathrm{v}}{\mathrm{A}-4}$
4 $\frac{2 \mathrm{v}}{\mathrm{A}-4}$
Explanation:
C As, $\mathrm{f}_{\text {net }}=0$, therefore, momentum is conserved. $\overrightarrow{\mathrm{P}}_{1}+\overrightarrow{\mathrm{P}}_{2}=0$ $\left|\overrightarrow{\mathrm{P}}_{1}\right|=\left|\overrightarrow{\mathrm{P}}_{2}\right|$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\mathrm{~m}(\mathrm{~A}-4) \mathrm{v}_{1}=4 \mathrm{mv}$ $\mathrm{v}_{1}=\frac{4 \mathrm{v}}{\mathrm{A}-4}$
AIIMS-2004
NUCLEAR PHYSICS
147720
A radioactive material decays by simultaneous emission of two particles with half-lives $1620 \mathrm{yr}$ and $810 \mathrm{yr}$ respectively. The time in years after which one-fourth of material remains, is-
1 $1080 \mathrm{yr}$
2 $2340 \mathrm{yr}$
3 $4860 \mathrm{yr}$
4 $3240 \mathrm{yr}$
Explanation:
A Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ The number of half lives (n) $=\frac{\text { timeof decay }}{\text { effectivehalf }- \text { life }}$ Relation between effective disintegration constant $(\lambda)$ and half life $\left(\mathrm{T}_{1 / 2}\right)$. $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\therefore \quad\lambda_{1}+\lambda_{2}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{1}}+\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{2}}$ Effective half life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}=\frac{1}{1620}+\frac{1}{810}$ $\frac{1}{\mathrm{~T}}=\frac{1+2}{1620}$ $\mathrm{~T}=540 \mathrm{yr}$ $\therefore \quad \mathrm{n}=\frac{\mathrm{T}}{540}$ $\therefore \quad \mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{t}}{540}=2$ $\text { So, } \quad \mathrm{t}=2 \times 540=1080 \mathrm{yr}$
BCECE-2015
NUCLEAR PHYSICS
147724
What is the disintegration constant of radon, if the number of its atoms diminishes by $18 \%$ in $24 \mathrm{~h}$ ?
1 $2.1 \times 10^{-3} \mathrm{~s}^{-1}$
2 $2.1 \times 10^{-5} \mathrm{~s}^{-1}$
3 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
4 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
Explanation:
D Undisintegrated part $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=(100-18) \%$ $=82 \%$ According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{82}{100}=\mathrm{e}^{-(24 \times 60 \times 60 \times \lambda)}$ $24 \times 60 \times 60 \times \lambda=\ln \left(\frac{100}{82}\right)$ $\lambda=2.2 \times 10^{-6} \mathrm{~s}^{-1}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
147711
Assertion: Cobalt-60 is useful in cancer therapy. Reason: Cobalt-60 is a source of $\gamma$-radiations capable of killing cancerous cells.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
A Cobalt-60 is a source of $\gamma$-radiation, which are used in curing cancer by destroying the cancerous cell. So, cobalt-60 is useful in cancer therapy.
AIIMS-2006
NUCLEAR PHYSICS
147718
In nucleus of mass number $A$, originally at rest, emits an $\alpha$-particle with speed $v$. The daughter nucleus recoils with a speed :
1 $\frac{2 \mathrm{v}}{\mathrm{A}+4}$
2 $\frac{4 \mathrm{v}}{\mathrm{A}+4}$
3 $\frac{4 \mathrm{v}}{\mathrm{A}-4}$
4 $\frac{2 \mathrm{v}}{\mathrm{A}-4}$
Explanation:
C As, $\mathrm{f}_{\text {net }}=0$, therefore, momentum is conserved. $\overrightarrow{\mathrm{P}}_{1}+\overrightarrow{\mathrm{P}}_{2}=0$ $\left|\overrightarrow{\mathrm{P}}_{1}\right|=\left|\overrightarrow{\mathrm{P}}_{2}\right|$ $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $\mathrm{~m}(\mathrm{~A}-4) \mathrm{v}_{1}=4 \mathrm{mv}$ $\mathrm{v}_{1}=\frac{4 \mathrm{v}}{\mathrm{A}-4}$
AIIMS-2004
NUCLEAR PHYSICS
147720
A radioactive material decays by simultaneous emission of two particles with half-lives $1620 \mathrm{yr}$ and $810 \mathrm{yr}$ respectively. The time in years after which one-fourth of material remains, is-
1 $1080 \mathrm{yr}$
2 $2340 \mathrm{yr}$
3 $4860 \mathrm{yr}$
4 $3240 \mathrm{yr}$
Explanation:
A Since, from Rutherford-Soddy law, the number of atoms left after half lives is given by - $\mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{n}}$ The number of half lives (n) $=\frac{\text { timeof decay }}{\text { effectivehalf }- \text { life }}$ Relation between effective disintegration constant $(\lambda)$ and half life $\left(\mathrm{T}_{1 / 2}\right)$. $\lambda=\frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ $\therefore \quad\lambda_{1}+\lambda_{2}=\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{1}}+\frac{\ln 2}{\left(\mathrm{~T}_{1 / 2}\right)_{2}}$ Effective half life, $\frac{1}{\mathrm{~T}}=\frac{1}{\mathrm{~T}_{1}}+\frac{1}{\mathrm{~T}_{2}}=\frac{1}{1620}+\frac{1}{810}$ $\frac{1}{\mathrm{~T}}=\frac{1+2}{1620}$ $\mathrm{~T}=540 \mathrm{yr}$ $\therefore \quad \mathrm{n}=\frac{\mathrm{T}}{540}$ $\therefore \quad \mathrm{N}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{\mathrm{t} / 540}$ $\frac{\mathrm{t}}{540}=2$ $\text { So, } \quad \mathrm{t}=2 \times 540=1080 \mathrm{yr}$
BCECE-2015
NUCLEAR PHYSICS
147724
What is the disintegration constant of radon, if the number of its atoms diminishes by $18 \%$ in $24 \mathrm{~h}$ ?
1 $2.1 \times 10^{-3} \mathrm{~s}^{-1}$
2 $2.1 \times 10^{-5} \mathrm{~s}^{-1}$
3 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
4 $2.2 \times 10^{-6} \mathrm{~s}^{-1}$
Explanation:
D Undisintegrated part $\left(\frac{\mathrm{N}}{\mathrm{N}_{\mathrm{o}}}\right)=(100-18) \%$ $=82 \%$ According to radioactivity decay law, $\mathrm{N}=\mathrm{N}_{\mathrm{o}}\left(\mathrm{e}^{-\lambda t}\right)$ $\frac{82}{100}=\mathrm{e}^{-(24 \times 60 \times 60 \times \lambda)}$ $24 \times 60 \times 60 \times \lambda=\ln \left(\frac{100}{82}\right)$ $\lambda=2.2 \times 10^{-6} \mathrm{~s}^{-1}$