The energy released by the fusion of \(1.0~\text{kg}\) of hydrogen deep within the Sun (the energy released in fusion \(26~\text{MeV}\)) is:
1. \(39.1495 \times 10^{26}~\text{MeV}\)
2. \(35.106 \times 10^{26}~\text{MeV}\)
3. \(33.106 \times 10^{26}~\text{MeV}\)
4. \(37.106 \times 10^{26}~\text{MeV}\)

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ${}_{}{}^{235}U$ to be about 200MeV.
1. \(5.041 \times 10^4 \mathrm{~kg}.\)
2. \(2.074 \times 10^4 \mathrm{~kg}.\)
3. \(3.076 \times 10^4 \mathrm{~kg}.\)
4. \(4.026 \times 10^4 \mathrm{~kg}.\)

Consider the \(\mathrm{(D\text–T)}\) reaction (deuterium-tritium fusion); \({ }_1^2 \mathrm{H}+{ }_1^3 \mathrm{H} \longrightarrow{ }_2^4 \mathrm{He}+\mathrm{n}.\) The energy released in \(\text{MeV}\) in this reaction from the data is:
[take \({m}\left({ }_1^2 \mathrm H\right)=2.014102~ \text{u},\) \({m}\left({ }_2^4 \mathrm{He}\right)=4.002603 ~ \text{u} ,\) \({m}(\mathrm{n})=1.00867 ~\text{u} ,\) and \({m}\left({ }_1^3 \mathrm{H}\right)=3.016049~ \text{u} \)]
1. \(17.59~\text{MeV}\)
2. \(18.01~\text{MeV}\)
3. \(20.03~\text{MeV}\)
4. \(19.68~\text{MeV}\)

Consider the fission of \(_{92}^{238}\mathrm{U}\) by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are \({}_{58}^{140}\mathrm{Ce}\) and \({}_{44}^{99}\mathrm{Ru}\)$$. What is \(Q\) for this fission process? The relevant atomic and particle masses are:
\(\mathrm m\left(_{92}^{238}\mathrm{U}\right)= 238.05079~\text{u}\)
\(\mathrm m\left(_{58}^{140}\mathrm{Ce}\right)= 139.90543~\text{u}\)
\(\mathrm m\left(_{44}^{99}\mathrm{Ru}\right)= 98.90594~\text{u}\)
1. \(303.037~\text{MeV}\)
2. \(205.981~\text{MeV}\)
3. \(312.210~\text{MeV}\)
4. \(231.007~\text{MeV}\)

A source contains two phosphorous radio nuclides ${}_{15}{}^{32}P$ ($T_{1/2}$ = 14.3d) and ${}_{15}{}^{33}P$ ($T_{1/2}$ = 25.3d). Initially, 10% of the decays come from ${}_{15}{}^{33}P$. How long one must wait until 90% do so?

1. 208.5 days
2. 200.2 days
3. 263.9 days
4. 211.7 days

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. The neutron separation energies of the nuclei \(_{20}^{41}\mathrm{Ca}\) is:
Given that:
\(\begin{aligned} & \mathrm{m}\left({ }_{20}^{40} \mathrm{C a}\right)=39.962591~ \text{u}\\ & \mathrm{m}\left({ }_{20}^{41} \mathrm{C a}\right)=40.962278 ~\text{u} \end{aligned}\)
1. \(7.657~\text{MeV}\)
2. \(8.363~\text{MeV}\)
3. \(9.037~\text{MeV}\)
4. \(9.861~\text{MeV}\)

What is the height of the potential barrier for a head-on collision of two deuterons? (Assume that they can be taken as hard spheres of radius 2.0 fm.)

1. 300 keV
2. 360 keV
3. 376 keV
4. 356 keV

A 1000 MW fission reactor consumes half of its fuel in 5.00 yr. How much ${}_{92}{}^{235}U$ did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of, ${}_{92}{}^{235}U$ and that this nuclide is consumed only by the fission process.

1. 4386 kg.
2. 3076 kg.
3. 4772 kg.
4. 8799 kg.

The fission properties of ${}_{94}{}^{239}Pu$ are very similar to those of ${}_{92}{}^{235}U$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ${}_{94}{}^{239}Pu$ undergo fission?

1. \(2.5\times 10^{25}\) MeV
2. \(4.5\times 10^{25}\) MeV
3. \(2.5\times 10^{26}\) MeV
4. \(4.5\times 10^{26}\) MeV