363684
The binding energy per nucleon of dueterium and helium nuclei are \(1.1\,MeV\) and \(7\,MeV\) respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
1 \(24.6\,MeV\)
2 \(25.1\,MeV\)
3 \(22.3\,MeV\)
4 \(23.6\,MeV\)
Explanation:
\({ }_{1} H^{2}+{ }_{1} H^{2} \rightarrow 2 H e^{4}+\Delta E\) The binding energy per nucleon of a deuterium \( = 1.1\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 1.1 = 4.4\,MeV\). The binding energy per nucleon of a helium nuclei \( = 7\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 7 = 28\,MeV\). Hence, energy released. \(\Delta E = (28 - 4.4)\,MeV = 23.6\,MeV.\)
PHXII13:NUCLEI
363685
Pick out the correct statement from the following:
1 Energy released per unit mass of the reactant is less in case of fusion reaction
2 Packing fraction may be positive or may be negative
3 \(P{u^{239}}\) is not suitable for a fission reaction
4 For stable nucleus, the specific binding energy is low
Explanation:
Packing fraction may be positive or negative.
KCET - 2010
PHXII13:NUCLEI
363686
Average binding energy per nucleon over a wide range is
1 \(8\,MeV\)
2 \(8.8\,MeV\)
3 \(5.6\,MeV\)
4 \(1.1\,MeV\)
Explanation:
Conceptual Question
PHXII13:NUCLEI
363687
The binding energy of two nuclei \({p^n}\) and \({Q^{2n}}\) are joule \(x\) and \(y\) joule respectively. If \(2x > y,\)then the energy change in the reaction \({p^n} + {p^n} = {Q^{2n}}\) will be
1 \(2x - y\)
2 \(2x + y\)
3 \(x + y\)
4 \(xy\)
Explanation:
Energy released \( = \) intial \(BE\) \( - \) final \(BE\) \( = 2x - y\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII13:NUCLEI
363684
The binding energy per nucleon of dueterium and helium nuclei are \(1.1\,MeV\) and \(7\,MeV\) respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
1 \(24.6\,MeV\)
2 \(25.1\,MeV\)
3 \(22.3\,MeV\)
4 \(23.6\,MeV\)
Explanation:
\({ }_{1} H^{2}+{ }_{1} H^{2} \rightarrow 2 H e^{4}+\Delta E\) The binding energy per nucleon of a deuterium \( = 1.1\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 1.1 = 4.4\,MeV\). The binding energy per nucleon of a helium nuclei \( = 7\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 7 = 28\,MeV\). Hence, energy released. \(\Delta E = (28 - 4.4)\,MeV = 23.6\,MeV.\)
PHXII13:NUCLEI
363685
Pick out the correct statement from the following:
1 Energy released per unit mass of the reactant is less in case of fusion reaction
2 Packing fraction may be positive or may be negative
3 \(P{u^{239}}\) is not suitable for a fission reaction
4 For stable nucleus, the specific binding energy is low
Explanation:
Packing fraction may be positive or negative.
KCET - 2010
PHXII13:NUCLEI
363686
Average binding energy per nucleon over a wide range is
1 \(8\,MeV\)
2 \(8.8\,MeV\)
3 \(5.6\,MeV\)
4 \(1.1\,MeV\)
Explanation:
Conceptual Question
PHXII13:NUCLEI
363687
The binding energy of two nuclei \({p^n}\) and \({Q^{2n}}\) are joule \(x\) and \(y\) joule respectively. If \(2x > y,\)then the energy change in the reaction \({p^n} + {p^n} = {Q^{2n}}\) will be
1 \(2x - y\)
2 \(2x + y\)
3 \(x + y\)
4 \(xy\)
Explanation:
Energy released \( = \) intial \(BE\) \( - \) final \(BE\) \( = 2x - y\)
363684
The binding energy per nucleon of dueterium and helium nuclei are \(1.1\,MeV\) and \(7\,MeV\) respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
1 \(24.6\,MeV\)
2 \(25.1\,MeV\)
3 \(22.3\,MeV\)
4 \(23.6\,MeV\)
Explanation:
\({ }_{1} H^{2}+{ }_{1} H^{2} \rightarrow 2 H e^{4}+\Delta E\) The binding energy per nucleon of a deuterium \( = 1.1\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 1.1 = 4.4\,MeV\). The binding energy per nucleon of a helium nuclei \( = 7\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 7 = 28\,MeV\). Hence, energy released. \(\Delta E = (28 - 4.4)\,MeV = 23.6\,MeV.\)
PHXII13:NUCLEI
363685
Pick out the correct statement from the following:
1 Energy released per unit mass of the reactant is less in case of fusion reaction
2 Packing fraction may be positive or may be negative
3 \(P{u^{239}}\) is not suitable for a fission reaction
4 For stable nucleus, the specific binding energy is low
Explanation:
Packing fraction may be positive or negative.
KCET - 2010
PHXII13:NUCLEI
363686
Average binding energy per nucleon over a wide range is
1 \(8\,MeV\)
2 \(8.8\,MeV\)
3 \(5.6\,MeV\)
4 \(1.1\,MeV\)
Explanation:
Conceptual Question
PHXII13:NUCLEI
363687
The binding energy of two nuclei \({p^n}\) and \({Q^{2n}}\) are joule \(x\) and \(y\) joule respectively. If \(2x > y,\)then the energy change in the reaction \({p^n} + {p^n} = {Q^{2n}}\) will be
1 \(2x - y\)
2 \(2x + y\)
3 \(x + y\)
4 \(xy\)
Explanation:
Energy released \( = \) intial \(BE\) \( - \) final \(BE\) \( = 2x - y\)
363684
The binding energy per nucleon of dueterium and helium nuclei are \(1.1\,MeV\) and \(7\,MeV\) respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is:
1 \(24.6\,MeV\)
2 \(25.1\,MeV\)
3 \(22.3\,MeV\)
4 \(23.6\,MeV\)
Explanation:
\({ }_{1} H^{2}+{ }_{1} H^{2} \rightarrow 2 H e^{4}+\Delta E\) The binding energy per nucleon of a deuterium \( = 1.1\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 1.1 = 4.4\,MeV\). The binding energy per nucleon of a helium nuclei \( = 7\,MeV\). \(\therefore\) Total binding energy \( = 4 \times 7 = 28\,MeV\). Hence, energy released. \(\Delta E = (28 - 4.4)\,MeV = 23.6\,MeV.\)
PHXII13:NUCLEI
363685
Pick out the correct statement from the following:
1 Energy released per unit mass of the reactant is less in case of fusion reaction
2 Packing fraction may be positive or may be negative
3 \(P{u^{239}}\) is not suitable for a fission reaction
4 For stable nucleus, the specific binding energy is low
Explanation:
Packing fraction may be positive or negative.
KCET - 2010
PHXII13:NUCLEI
363686
Average binding energy per nucleon over a wide range is
1 \(8\,MeV\)
2 \(8.8\,MeV\)
3 \(5.6\,MeV\)
4 \(1.1\,MeV\)
Explanation:
Conceptual Question
PHXII13:NUCLEI
363687
The binding energy of two nuclei \({p^n}\) and \({Q^{2n}}\) are joule \(x\) and \(y\) joule respectively. If \(2x > y,\)then the energy change in the reaction \({p^n} + {p^n} = {Q^{2n}}\) will be
1 \(2x - y\)
2 \(2x + y\)
3 \(x + y\)
4 \(xy\)
Explanation:
Energy released \( = \) intial \(BE\) \( - \) final \(BE\) \( = 2x - y\)