- 1(current)
Q. No.(Avogadro Number \(= 6\times 10^{23}\) per mole)
1. \(220\)
2. \(228\)
3. \(210\)
4. \(230\)
| List-I | List-II | ||
| \(\mathrm{A.}\) | \({ }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_0^1 \mathrm{n}\) | \(\mathrm{I.}\) | Chemical reaction |
| \(\mathrm{B.}\) | \(2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}\) | \(\mathrm{II.}\) | Fusion with \(+\text{ve}~Q \) value |
| \(\mathrm{C.}\) | \({ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n}\) | \(\mathrm{III.}\) | Fission |
| \(\mathrm{D.}\) | \({ }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}\) | \(\mathrm{IV.}\) | Fusion with \(-\text{ve}~Q \) value |
| 1. | \(\mathrm{A\text-III, B\text-I, C\text-II, D\text-IV}\) |
| 2. | \(\mathrm{A\text-II, B\text-I, C\text-III, D\text-IV}\) |
| 3. | \(\mathrm{A\text-II, B\text-I, C\text-IV, D\text-III}\) |
| 4. | \(\mathrm{A\text-III, B\text-I, C\text-IV, D\text-II}\) |
| \(\mathrm{(A)}\) | The angular momentum of orbiting electrons is the integral multiplication of \(h\) |
| \(\mathrm{(B)}\) | The nuclear force does not follow the inverse square law. |
| \(\mathrm{(C)}\) | The nuclear force is dependent on the spin quantum number. |
| \(\mathrm{(D)}\) | The nuclear force is independent of the nature of nucleons. |
| \(\mathrm{(E)}\) | The stability of the nucleus depends upon the packing fraction. |
| 1. | \(\mathrm{A, B, C, D}\) are correct | 2. | \(\mathrm{A, B, C, E}\) are correct |
| 3. | \(\mathrm{A, C, D, E}\) are correct | 4. | \(\mathrm{B, C, D, E}\) are correct |
The masses of lithium-\(7\) \(({ }_3^7 \mathrm{Li}),\) helium-\(4\) \(({ }_2^4 \mathrm{He}),\) and hydrogen-\(1\) \(({ }_1^1 \mathrm{H})\) are \(7.0160~\text{u},\) \(4.0026~\text{u},\) and \(1.0079~\text{u},\) respectively. When \(20~\text{g}\) of lithium-\(7\) undergoes proton capture according to the reaction \({}^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{H}\rightarrow 2\,{}^{4}_{2}\mathrm{He},\) calculate the total energy released in kWh. Given: \(1~\text{u}=931.5~\text{MeV/c}^2,\) \(1~\text{MeV}=1.602 \times10^{-13}~\text{J},\) Avogadro's number = \(6.022 \times10^{23}\) mol⁻¹, and \(1\) kWh \(3.6\times 10^{6}\) J.
1. \( 1.33 \times 10^6 \)
2. \( 8 \times 10^6 \)
3. \( 6.82 \times 10^5 \)
4. \( 4.5 \times 10^5 \)
- 1(current)
Q. No.
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