Q. No.The ratio of the mass densities of the nuclei \({ }^{40} \mathrm{Ca}\) and \({ }^{16} \mathrm{O}\) is close to:
1.
\(0.1\)
2.
\(2\)
3.
\(5\)
4.
\(1\)
In a reactor, \(2\) kg of \({ }_{92} \mathrm{U}^{235}\) fuel is fully used up in \(30\) days. The energy released per fission is \(200\) MeV. Given that the Avogadro number, \(\mathrm{N}=6.023 \times 10^{26} \) per kilo mole and \(1~ \mathrm{eV}=1.6 \times 10^{-19}~\text{J}\). The power output of the reactor is close to:
1. \(125 ~\text{MW}\)
2. \(60~\text{MW}\)
3. \(35 ~\text{MW}\)
4. \(54 ~\text{MW}\)
The radius \(R\) of a nucleus of mass number \(A\) can be estimated by the formula \({R}=\left(1.3 \times 10^{-15}\right) A^{1 / 3} ~\text{m}\) , It follows that the mass density of a nucleus is of the order of: \(\left(M_{\text {propt. }}=M_{\text {neut. }}=1.67 \times 10^{-27} ~\text{kg}\right)\)
1. \( 10^{10}~ \text{kg}\text{m}^{-3} \)
2. \( 10^{24} ~\text{kg} \text{m}^{-3} \)
3. \( 10^{17} ~\text{kg} \text{m}^{-3} \)
4. \( 10^{3} ~\text{kg} \text{m}^{-3} \)
An \(\mathrm{X} \text-\)ray beam has a wavelength of \(10 ~\mathring{A}.\) A fictitious particle has the same energy as that of an \(\mathrm{X} \text-\)ray photon. If the mass of this particle is expressed as \(m=\dfrac{xh}{3}~\text{kg}, \) where \(h\) is Planck’s constant, what is the value of \(x\)?
| 1. | \(15\) | 2. | \(10\) |
| 3. | \(20\) | 4. | \(25\) |
Given the following particle masses:
\(m_p=1.0072~\text{u}\) (proton)
\(m_n=1.0087~\text{u}\) (neutron)
\(m_e=0.000548~\text{u}\) (electron)
\(m_\nu=0~\text{u}\) (antineutrino)
\(m_d=2.0141~\text{u}\) (deuteron)
Which of the following processes is allowed, considering the conservation of energy and momentum?
| 1. | \(n+p \rightarrow d+\gamma\) |
| 2. | \(e^{+}+e^{-} \rightarrow \gamma\) |
| 3. | \(n+n\rightarrow \text{}\) deuterium atom (electron bound to the nucleus) |
| 4. | \(p \rightarrow n+e^{+}+\nu\) |
You are given that mass of \({ }_a^7 \mathrm{Li}=7.0160 ~\text{u}\) Mass of \({ }_2^4 \mathrm{He}=4.0026 ~\text{u}\) and Mass of \({ }_1^1 \mathrm{H}=1.0079 ~\text{u}\)
When \(20~\text{g}\) of \({ }_a^7 \mathrm{Li}\) is converted into \({ }_2^4 \mathrm{He}\) By proton capture, the energy liberated, (in kWh), is:
[Mass of nucleon = \(1~\text{GeV/c}^2\)]
1. \( 1.33 \times 10^6 \)
2. \( 8 \times 10^6 \)
3. \( 6.82 \times 10^5 \)
4. \( 4.5 \times 10^5 \)
Given that the masses of a proton, a neutron, and the nucleus of \({ }_{50}^{120} \mathrm{Sn}\) are \(1.00783~\text{u},\) \(1.00867~\text{u},\) and \(119.902199~ \text{u},\) respectively. The binding energy per nucleon of the tin nucleus is: \((1~\text{u}=931~\text{Mev})\)
| 1. | \(9~\text{MeV}\) | 2. | \(8.5~\text{MeV}\) |
| 3. | \(8.0~\text{MeV}\) | 4. | \(7.5~\text{MeV}\) |
1. 0.02 min –1
2. 2.7 min –1
3. 0.063 min –1
4. 6.3 min–1
1. \(4: 3 \)
2. \(\left(\frac{3}{4}\right)^{\frac{1}{3}} \)
3. \(1: 1 \)
4. \(\left(\frac{4}{3}\right)^{\frac{1}{3}}\)
\({ }_{1}^{2} \mathrm{X}+{ }_{1}^{2} \mathrm{X}={ }_{2}^{4} \mathrm{Y}\)
The binding energies per nucleon for \({ }_{1}^{2} \mathrm{X} \text { and }{ }_{2}^{4} \mathrm{Y}\) are \(1.1~\text{MeV}\) and \(7.6~\text{MeV}\) respectively. The energy released in this process is:
1. \(26~\text{MeV}\)
2. \(34~\text{MeV}\)
3. \(42~\text{MeV}\)
4. \(24~\text{MeV}\)
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