The energy equivalent of one atomic mass unit is:
1. $$1.6\times 10^{-19}~\text{J}$$
2. $$6.02\times 10^{23}~\text{J}$$
3. $$931~\text{MeV}$$
4. $$9.31~\text{MeV}$$

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A nuclear reaction along with the masses of the particle taking part in it is as follows;
$$~~A ~~~~+~~~ B~~~~~ \rightarrow~~~~C ~~+~~~~ D~~~~~ ~~Q~ MeV\\ \small{1.002~~~~~~~~ 1.004 ~~~~~~~~~~~~~1.001~~~~~~~1.003}\\ \small{amu~~~~~~~~~~amu~~~~~~~~~~~~~~amu~~~~~~~~~amu} ~~$$
The energy $$Q$$ liberated in the reaction is:
1. $$1.234$$ MeV
2. $$0.931$$ MeV
3. $$0.465$$ MeV
4. $$1.862$$ MeV

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Determine the energy released in the process:
$${}_{1}^{2}\mathrm{H}+ {}_{1}^{2}\mathrm{H}\rightarrow {}_{2}^{4}\mathrm{He}+Q$$
Given: $$M\left({}_{1}^{2}\mathrm{H}\right)= 2.01471~\text{amu}, M\left({}_{2}^{4}\mathrm{He}\right)= 4.00388~\text{amu}$$
1. $$3.79$$ MeV
2. $$13.79$$ MeV
3. $$0.79$$ MeV
4. $$23.79$$ MeV
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If an electron and a positron annihilate, then the energy released is:
1. $$3.2\times 10^{-13}~\text{J}$$
2. $$1.6\times 10^{-13}~\text{J}$$
3. $$4.8\times 10^{-13}~\text{J}$$
4. $$6.4\times 10^{-13}~\text{J}$$

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The energy required in $$\text{MeV/c}^2$$ to separate $${ }_8^{16} \mathrm{O}$$ into its constituents is:
(Given: mass defect for $${ }_8^{16} \mathrm{O}=0.13691~ \text{amu}$$)
1. $$127.5$$
2. $$120.0$$
3. $$222.0$$
4. $$119.0$$

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A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in the fusion reaction is $$0.02866$$ u. The energy liberated per nucleon is: (Given $$1$$ u = $$931$$ MeV)
 1 $$26.7$$ MeV 2 $$6.675$$ MeV 3 $$13.35$$ MeV 4 $$2.67$$ MeV
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AIPMT - 2013
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The energy equivalent of $$0.5$$ g of a substance is:
1. $$4.5\times10^{13}$$ J
2. $$1.5\times10^{13}$$ J
3. $$0.5\times10^{13}$$ J
4. $$4.5\times10^{16}$$ J

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NEET - 2020
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If a proton and anti-proton come close to each other and annihilate, how much energy will be released?

 1 $$1.5 \times10^{-10}~\text{J}$$ 2 $$3 \times10^{-10}~\text{J}$$ 3 $$4.5 \times10^{-10}~\text{J}$$ 4 None of these
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Calculate the $$Q\text-$$value of the nuclear reaction:
$$2~{ }_{6}^{12} \mathrm{C}\rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_2^4 \mathrm{He}$$
The following data are given:
$$m({ }_{6}^{12} \mathrm{C})=12.000000~\text{amu}$$
$$m({ }_{10}^{20} \mathrm{Ne})=19.992439~\text{amu}$$
$$m({ }_{2}^{4} \mathrm{He})=4.002603~\text{amu}$$
1. $$3.16~\text{MeV}$$
2. $$5.25~\text{MeV}$$
3. $$3.91~\text{MeV}$$
4. $$4.65~\text{MeV}$$

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The rest energy of an electron is:
 1 $$510$$ KeV 2 $$931$$ KeV 3 $$510$$ MeV 4 $$931$$ MeV
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