NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII13:NUCLEI
363637
The energy released by the fission of \(1\,g\) of \({ }^{235} U\) in joule, given the energy released per fission is \(200\,MeV\) (Avagadro's number \(=6.023 \times 10^{23}\))
1 \(8.202 \times 10^{12}\)
2 \(8.202 \times 10^{8}\)
3 \(8.202 \times 10^{10}\)
4 \(8.202 \times 10^{14}\)
Explanation:
When ' \(x\) ' gram of nuclear fuel of mass no. \(A\) undergoes fission completely, then the number of fissions \(n=\) same as number of atoms. \(\Rightarrow n=\dfrac{N x}{A}\) where \(N\) is Avagadro no. \(\Rightarrow n=\dfrac{6.023 \times 10^{23} \times 10^{-3}}{235 \times 10^{-3}}=\dfrac{6.023 \times 10^{23}}{235}\) \(\Rightarrow Q=n E=\dfrac{6.023 \times 10^{23}}{235} \times 200 \times 1.6 \times 10^{-13}\) \( = 8.202 \times {10^{10}}\;J.\)
PHXII13:NUCLEI
363638
The energy in \(MeV\) is released due to transformation of \(1\;kg\) mass completely into energy:
363639
A nucleus with mass number 220 initially at rest emits an alpha particle. If the \(Q\) value of reaction is \(5.5\,MeV\), calculate the value of kinetic energy of alpha particle.
1 \(7.4\,MeV\)
2 \(4.5\,MeV\)
3 \(6.5\,MeV\)
4 \(5.4\,MeV\)
Explanation:
Required \(\left(E_{K}\right)_{\alpha}\) is \(\Rightarrow\left(E_{K}\right)_{\alpha}=Q\left(\dfrac{A-4}{A}\right)\) \(\Rightarrow\left(E_{K}\right)_{\alpha}=5.5 \times \dfrac{216}{220}\) \( = 5.4\,MeV\). Correct option is (4).
KCET - 2023
PHXII13:NUCLEI
363640
When three \(\alpha\)-particles are combined to form a \({C^{12}}\) nucleus, the mass defect is (Atomic mass of \(_2H{e^4}\,\,4.002603{\mkern 1mu} \,u)\)
1 \(0.007809\,u\)
2 \(0.002603\,u\)
3 \(4.002603\,u\)
4 \(0.5\,u\)
Explanation:
The given reaction is \(3{\,_2}H{e^4}{ \to _6}{C^{12}}\) The mass defect for this reaction is \(\Delta M = \left[ {3{M_{He}} - {M_C}} \right]\) \(=[3(4.002603\, u)-12\, u]\) \(=12.007809\, u-12 u=0.007809\, u\)
363637
The energy released by the fission of \(1\,g\) of \({ }^{235} U\) in joule, given the energy released per fission is \(200\,MeV\) (Avagadro's number \(=6.023 \times 10^{23}\))
1 \(8.202 \times 10^{12}\)
2 \(8.202 \times 10^{8}\)
3 \(8.202 \times 10^{10}\)
4 \(8.202 \times 10^{14}\)
Explanation:
When ' \(x\) ' gram of nuclear fuel of mass no. \(A\) undergoes fission completely, then the number of fissions \(n=\) same as number of atoms. \(\Rightarrow n=\dfrac{N x}{A}\) where \(N\) is Avagadro no. \(\Rightarrow n=\dfrac{6.023 \times 10^{23} \times 10^{-3}}{235 \times 10^{-3}}=\dfrac{6.023 \times 10^{23}}{235}\) \(\Rightarrow Q=n E=\dfrac{6.023 \times 10^{23}}{235} \times 200 \times 1.6 \times 10^{-13}\) \( = 8.202 \times {10^{10}}\;J.\)
PHXII13:NUCLEI
363638
The energy in \(MeV\) is released due to transformation of \(1\;kg\) mass completely into energy:
363639
A nucleus with mass number 220 initially at rest emits an alpha particle. If the \(Q\) value of reaction is \(5.5\,MeV\), calculate the value of kinetic energy of alpha particle.
1 \(7.4\,MeV\)
2 \(4.5\,MeV\)
3 \(6.5\,MeV\)
4 \(5.4\,MeV\)
Explanation:
Required \(\left(E_{K}\right)_{\alpha}\) is \(\Rightarrow\left(E_{K}\right)_{\alpha}=Q\left(\dfrac{A-4}{A}\right)\) \(\Rightarrow\left(E_{K}\right)_{\alpha}=5.5 \times \dfrac{216}{220}\) \( = 5.4\,MeV\). Correct option is (4).
KCET - 2023
PHXII13:NUCLEI
363640
When three \(\alpha\)-particles are combined to form a \({C^{12}}\) nucleus, the mass defect is (Atomic mass of \(_2H{e^4}\,\,4.002603{\mkern 1mu} \,u)\)
1 \(0.007809\,u\)
2 \(0.002603\,u\)
3 \(4.002603\,u\)
4 \(0.5\,u\)
Explanation:
The given reaction is \(3{\,_2}H{e^4}{ \to _6}{C^{12}}\) The mass defect for this reaction is \(\Delta M = \left[ {3{M_{He}} - {M_C}} \right]\) \(=[3(4.002603\, u)-12\, u]\) \(=12.007809\, u-12 u=0.007809\, u\)
363637
The energy released by the fission of \(1\,g\) of \({ }^{235} U\) in joule, given the energy released per fission is \(200\,MeV\) (Avagadro's number \(=6.023 \times 10^{23}\))
1 \(8.202 \times 10^{12}\)
2 \(8.202 \times 10^{8}\)
3 \(8.202 \times 10^{10}\)
4 \(8.202 \times 10^{14}\)
Explanation:
When ' \(x\) ' gram of nuclear fuel of mass no. \(A\) undergoes fission completely, then the number of fissions \(n=\) same as number of atoms. \(\Rightarrow n=\dfrac{N x}{A}\) where \(N\) is Avagadro no. \(\Rightarrow n=\dfrac{6.023 \times 10^{23} \times 10^{-3}}{235 \times 10^{-3}}=\dfrac{6.023 \times 10^{23}}{235}\) \(\Rightarrow Q=n E=\dfrac{6.023 \times 10^{23}}{235} \times 200 \times 1.6 \times 10^{-13}\) \( = 8.202 \times {10^{10}}\;J.\)
PHXII13:NUCLEI
363638
The energy in \(MeV\) is released due to transformation of \(1\;kg\) mass completely into energy:
363639
A nucleus with mass number 220 initially at rest emits an alpha particle. If the \(Q\) value of reaction is \(5.5\,MeV\), calculate the value of kinetic energy of alpha particle.
1 \(7.4\,MeV\)
2 \(4.5\,MeV\)
3 \(6.5\,MeV\)
4 \(5.4\,MeV\)
Explanation:
Required \(\left(E_{K}\right)_{\alpha}\) is \(\Rightarrow\left(E_{K}\right)_{\alpha}=Q\left(\dfrac{A-4}{A}\right)\) \(\Rightarrow\left(E_{K}\right)_{\alpha}=5.5 \times \dfrac{216}{220}\) \( = 5.4\,MeV\). Correct option is (4).
KCET - 2023
PHXII13:NUCLEI
363640
When three \(\alpha\)-particles are combined to form a \({C^{12}}\) nucleus, the mass defect is (Atomic mass of \(_2H{e^4}\,\,4.002603{\mkern 1mu} \,u)\)
1 \(0.007809\,u\)
2 \(0.002603\,u\)
3 \(4.002603\,u\)
4 \(0.5\,u\)
Explanation:
The given reaction is \(3{\,_2}H{e^4}{ \to _6}{C^{12}}\) The mass defect for this reaction is \(\Delta M = \left[ {3{M_{He}} - {M_C}} \right]\) \(=[3(4.002603\, u)-12\, u]\) \(=12.007809\, u-12 u=0.007809\, u\)
363637
The energy released by the fission of \(1\,g\) of \({ }^{235} U\) in joule, given the energy released per fission is \(200\,MeV\) (Avagadro's number \(=6.023 \times 10^{23}\))
1 \(8.202 \times 10^{12}\)
2 \(8.202 \times 10^{8}\)
3 \(8.202 \times 10^{10}\)
4 \(8.202 \times 10^{14}\)
Explanation:
When ' \(x\) ' gram of nuclear fuel of mass no. \(A\) undergoes fission completely, then the number of fissions \(n=\) same as number of atoms. \(\Rightarrow n=\dfrac{N x}{A}\) where \(N\) is Avagadro no. \(\Rightarrow n=\dfrac{6.023 \times 10^{23} \times 10^{-3}}{235 \times 10^{-3}}=\dfrac{6.023 \times 10^{23}}{235}\) \(\Rightarrow Q=n E=\dfrac{6.023 \times 10^{23}}{235} \times 200 \times 1.6 \times 10^{-13}\) \( = 8.202 \times {10^{10}}\;J.\)
PHXII13:NUCLEI
363638
The energy in \(MeV\) is released due to transformation of \(1\;kg\) mass completely into energy:
363639
A nucleus with mass number 220 initially at rest emits an alpha particle. If the \(Q\) value of reaction is \(5.5\,MeV\), calculate the value of kinetic energy of alpha particle.
1 \(7.4\,MeV\)
2 \(4.5\,MeV\)
3 \(6.5\,MeV\)
4 \(5.4\,MeV\)
Explanation:
Required \(\left(E_{K}\right)_{\alpha}\) is \(\Rightarrow\left(E_{K}\right)_{\alpha}=Q\left(\dfrac{A-4}{A}\right)\) \(\Rightarrow\left(E_{K}\right)_{\alpha}=5.5 \times \dfrac{216}{220}\) \( = 5.4\,MeV\). Correct option is (4).
KCET - 2023
PHXII13:NUCLEI
363640
When three \(\alpha\)-particles are combined to form a \({C^{12}}\) nucleus, the mass defect is (Atomic mass of \(_2H{e^4}\,\,4.002603{\mkern 1mu} \,u)\)
1 \(0.007809\,u\)
2 \(0.002603\,u\)
3 \(4.002603\,u\)
4 \(0.5\,u\)
Explanation:
The given reaction is \(3{\,_2}H{e^4}{ \to _6}{C^{12}}\) The mass defect for this reaction is \(\Delta M = \left[ {3{M_{He}} - {M_C}} \right]\) \(=[3(4.002603\, u)-12\, u]\) \(=12.007809\, u-12 u=0.007809\, u\)
