- 1(current)
Q. No.| 1. | Binding energy per nucleon is practically constant for nuclei with mass numbers between \(30\) and \(170\). |
| 2. | Binding energy per nucleon is maximum for \(_{56}\mathrm{Fe}\) (equal to \(8.75~\text{MeV}\)). |
| 3. | Binding energy per nucleon for \(_{6}\mathrm{Li}\) is lower compared to \(_{4}\mathrm{He}\). |
| 4. | Higher the binding energy per nucleon, the more unstable is the nucleus. |
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| 1. | \(E_1\): total binding energy of initial nuclei |
| 2. | \(E_2\): total binding energy of final nuclei |
| 3. | \(A_1\): total number of nucleons of initial nuclei |
| 4. | \(A_2\): total number of nucleons of final nuclei |
1. \(E_1>E_2\)
2. \(E_2>E_1\)
3. \(E_1=E_2\)
4. \(\dfrac{E_1}{A_1}=\dfrac{E_2}{A_2}\)
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| 1. | \( \mathrm{F}_{\mathrm{pp}}<\mathrm{F}_{\mathrm{pn}}<\mathrm{F}_{\mathrm{nn}} \) | 2. | \( \mathrm{F}_{\mathrm{pn}}>\mathrm{F}_{\mathrm{pp}}>\mathrm{F}_{\mathrm{nn}} \) |
| 3. | \( \mathrm{F}_{\mathrm{pp}}>\mathrm{F}_{\mathrm{pn}}>\mathrm{F}_{\mathrm{nn}} \) | 4. | \(\mathrm{F}_{\mathrm{pp}}=\mathrm{F}_{\mathrm{pn}}=\mathrm{F}_{\mathrm{nn}}\) |
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| Assertion (A): | Binding energy per nucleon for nuclei (atomic number \(30\) to \(107\)) is independent of atomic number. |
| Reason (R): | Nuclear force is short-range force. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | Both (A) and (R) are False. |
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| 1. | is only attractive force. |
| 2. | is only repulsive force. |
| 3. | maybe attractive or repulsive in nature depending on the distance. |
| 4. | is a central force. |
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1. An electric force
2. Weak nuclear force
3. Strong nuclear force
4. Gravitational force
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\(^{298}_{94}X \rightarrow ^{294}_{92}Y + { ^{4}_{2}\alpha} + Q\text-\text {value}\),
where the binding energy per nucleon of \(X,Y \) and \(\alpha\) are denoted by \(a, b \) and \(c\) respectively.
The expression for the \(Q\)-value is:
1. \((294 b +4c - 298 a)\)
2. \((92 b +2c - 94 a)\)
3. \((294 b +4c + 298 a)\)
4. \((92 b +2c + 94 a)\)
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We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}=238.05079~\text{u},{ }_2^4 \mathrm{He}=4.00260~\text{u} \\ { }_{90}^{234} \mathrm{Th}=234.04363~\text{u},{ }_1^1 \mathrm{H}=1.00783~\text{u}\\ { }_{91}^{237} \mathrm{~Pa}=237.05121~\text{u} \)
Here the symbol \(\mathrm{Pa}\) is for the element protactinium \((Z=91)\).
The energy released during the alpha decay of \({}^{238}_{92}\mathrm{U}\) is:
1. \(6.14~\text{MeV}\)
2. \(7.68~\text{MeV}\)
3. \(4.25~\text{MeV}\)
4. \(5.01~\text{MeV}\)
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We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}=238.05079~\text{u},{ }_2^4 \mathrm{He}=4.00260~\text{u} \\ { }_{90}^{234} \mathrm{Th}=234.04363~\text{u},{ }_1^1 \mathrm{H}=1.00783~\text{u}\\ { }_{91}^{237} \mathrm{~Pa}=237.05121~\text{u} \)
Here the symbol Pa is for the element protactinium \((Z=91)\).
Then:
| 1. | \({}_{92}^{238}\mathrm{U}\) can not spontaneously emit a proton. |
| 2. | \({}_{92}^{238}\mathrm{U}\) can spontaneously emit a proton. |
| 3. | The \(Q\text-\)value of the process is negative. |
| 4. | Both (1) and (3) are correct. |
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- 1(current)
Q. No.
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