What is the binding energy (in MeV) of a nitrogen nucleus ${}_7{}^{14}N$?

$\mathrm{Given},$
${\mathrm{m}}_{\mathrm{p}} = 1.007825 \mathrm{u}$
${\mathrm{m}}_{\mathrm{n}} = 1.008665 \mathrm{u}$
$\mathrm{m}\left({}_7{}^{14}\mathrm{N}\right) = 14.003074 u$

1. 102.7 MeV.
2. 100.7 MeV.
3. 104.7 MeV.
4. 108.7 MeV.

A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to 3.125% of its original value?

1. T years.
2. 4T years.
3. 3T years.
4. 5T years.

What is the amount of ${}_{27}{}^{60}Co$ necessary to provide a radioactive source of 8.0 mCi strength? The half-life of ${}_{27}{}^{60}Co$ is 5.3 years.

1. $8.109\times {10}^{-6} g$
2. $7.106\times {10}^{-6} g$ 
3. $7.105\times {10}^{-5} g$
4. $8.107\times {10}^{-5} g$

A given coin has a mass of \(3.0~\text g.\) The nuclear energy required to separate all the neutrons and protons from each other will be:
(for simplicity assume that the coin is entirely made of \({}^{63}_{29}\mathrm{Cu}\) atoms of mass \(62.92960~\text u,\) the mass of proton \(m_p=1.00783~\text u,\) and the mass of neutron \(m_n=1.00867 ~\text u\))
1. \(2.5296\times10^{12}~\text{MeV}\)
2. \(1.581\times10^{25}~\text{MeV}\)  
3. \(3.1223\times10^{20}~\text{MeV}\)
4. \(931.02\times10^{19}~\text{MeV}\)

The amount of ${}_{27}{}^{60}Co$ necessary to provide a radioactive source of 8.0 mCi strength is:

(The half-life of ${}_{27}{}^{60}Co$ is 5.3 years)

1.  \(6.3\times10^{-6}\) g
2. \(7.1\times10^{-6}\) g
3. \(5.7\times10^{-6}\) g
4. \(6.9\times10^{-6}\) g

The half-life of ${}_{38}{}^{90}Sr$ is 28 years. What is the disintegration rate of 15 mg of this isotope?
1. \(9.64 \times 10^{10}~\mathrm{atoms} / \mathrm{s}\)
2. \(11.12 \times 10^{11}~\mathrm{atoms} / \mathrm{s}\)
3. \(7.87 \times 10^{10}~\mathrm{atoms}/ \mathrm{s}\)
4. \(10.04 \times 10^{11}~\mathrm{atoms}/ \mathrm{s}\)

The approximately nuclear radii ratio of the gold isotope \(_{79}^{197}\textrm{Au}\) and the silver isotope \(_{47}^{107}\textrm{Au}\) is:
1. \(1: 1.23\)
2. \(1 : 1.32\)
3. \(1.01 : 1\)
4. \(1.22 : 1\)

The radionuclide \(^{11}_{6}C\) decays according to \(^{11}_{6}C \rightarrow ~^{11}_{5}B+e^{+}+\nu\): \(\left(T_{\frac{1}{2}}=20.3~\text{min}\right)\)
The maximum energy of the emitted position is \(0.960~\text{MeV}\).
Given the mass values:
\(m\left(_{6}^{11}C\right) = 11.011434~\text{u}~\text{and}~ m\left(_{6}^{11}B\right) = 11.009305~\text{u},\)
The value of \(Q\) is:
1. \(0.313~\text{MeV}\)
2. \(0.962~\text{MeV}\)
3. \(0.414~\text{MeV}\)
4. \(0.132~\text{MeV}\)

The nucleus ${}_{10}{}^{23}\mathrm{Ne}$ decays by β– emission. What is the maximum kinetic energy of the electrons emitted? Given that:

m (${}_{10}{}^{23}Ne$) = 22.994466 u

m (${}_{11}{}^{23}Na$) = 22.989770 u.

1. 4.201 MeV
2. 3.791 MeV
3. 4.374 MeV
4. 3.851 MeV

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