A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit ( u ) is defined as 1 12 the mass of the carbon-12 atom. Calculate: (a) the mass defect in kilograms (Mass of a proton = 1 . 007276 u , mass of a neutron = 1 . 008665 u , mass of an electron = 0 . 000548 u .)

Given that Mass of a proton = 1 . 007276 u Mass of a neutron = 1 . 008665 u Mass of an electron = 0 . 000548 u Mass defect in u = 6 × ( 1 . 007 276 + 1 . 008 665 + 0 . 000 548 ) – 12 . 000 = 0 . 098 934 u Mass defect in k g = 0 . 098 934 × 1 . 660 × 10 - 27 = 1 . 64 × 10 - 28 k g

A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit ( u ) is defined as 1 12 the mass of the carbon-12 atom. Calculate: (b) the binding energy (Mass of a proton = 1 . 007276 u , mass of a neutron = 1 . 008665 u , mass of an electron = 0 . 000548 u .)

Given that Mass of a proton = 1 . 007276 u Mass of a neutron = 1 . 008665 u Mass of an electron = 0 . 000548 u Mass defect in u = 6 × ( 1 . 007 276 + 1 . 008 665 + 0 . 000 548 ) – 12 . 000 = 0 . 098 934 u Mass defect in k g = 0 . 098 934 × 1 . 660 × 10 - 27 = 1 . 64 × 10 - 28 k g The binding energy= ∆ m × c 2 = 1 . 64 × 10 - 29 × ( 3 . 0 × 10 8 ) 2 = 1 . 48 × 10 - 11 J

A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit ( u ) is defined as 1 12 the mass of the carbon-12 atom. Calculate: (c) the binding energy per nucleon. (Mass of a proton = 1 . 007276 u , mass of a neutron = 1 . 008665 u , mass of an electron = 0 . 000548 u .)

Given that Mass of a proton = 1 . 007276 u Mass of a neutron = 1 . 008665 u Mass of an electron = 0 . 000548 u Mass defect in u = 6 × ( 1 . 007 276 + 1 . 008 665 + 0 . 000 548 ) – 12 . 000 = 0 . 098 934 u Mass defect in k g = 0 . 098 934 × 1 . 660 × 10 - 27 = 1 . 64 × 10 - 28 k g The binding energy= ∆ m × c 2 = 1 . 64 × 10 - 29 × ( 3 . 0 × 10 8 ) 2 = 1 . 48 × 10 - 11 J Number of nucleons = 12 So, Binding energy per nucleon is 1 . 48 × 10 - 11 12 = 1 . 23 × 10 - 12 J

The fusion reaction that holds most promise for the generation of electricity is the fusion of tritium H 1 3 and deuterium H 1 2 . The following equation shows the process: H 1 3 + H 1 2 → He 2 4 + H 1 1 Calculate: (a) the change in mass in the reaction (Mass of H 1 3 = 3 . 015500 u , Mass of H 1 2 = 2 . 013553 u , Mass of He 2 4 = 4 . 00150 u , Mass of H 1 1 = 1 . 007276 u . )

The change in mass in the reaction is equal to decrease in mass. So, Decrease in mass = 3 . 015500 + 2 . 013553 – 4 . 001506 – 1 . 007276 = 0 . 020271 u = 0 . 020271 × 1 . 660 × 10 - 27 k g = 3 . 365 × 10 - 29 k g

The fusion reaction that holds most promise for the generation of electricity is the fusion of tritium H 1 3 and deuterium H 1 2 . The following equation shows the process: H 1 3 + H 1 2 → He 2 4 + H 1 1 Calculate: (b) the energy released in the reaction (Mass of H 1 3 = 3 . 015500 u , Mass of H 1 2 = 2 . 013553 u , Mass of He 2 4 = 4 . 00150 u , Mass of H 1 1 = 1 . 007276 u . )

The change in mass in the reaction is equal to decrease in mass. So, Decrease in mass = 3 . 015500 + 2 . 013553 – 4 . 001506 – 1 . 007276 = 0 . 020271 u = 0 . 020271 × 1 . 660 × 10 - 27 k g = 3 . 365 × 10 - 29 k g The energy released in the reaction = m c 2 = = 3 . 365 × 10 - 29 × ( 3 . 0 × 10 8 ) 2 = 3 . 028 × 10 - 12 J

MEDIUM

AS and A Level

IMPORTANT

Earn 100

An antiproton is identical to a proton except that it has negative charge. When a proton and an antiproton collide, they are annihilated and two photons are formed. In annihilation, all the mass of the particles is converted into energy. Calculate the energy released if $1$ mole of protons and $1$ mole of antiprotons were annihilated by this process. (Mass of a proton $=$ mass of an antiproton $= 1.67 \times 10^{- 27} kg$ . )

Important Questions on Nuclear Physics

EASY

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 4

Calculate the mass that would be annihilated to release $1 \mathrm{~J}$ of energy.

EASY

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 5

In a nuclear reactor, the mass converted to energy takes place at a rate of

70 μg s^{- 1}

. Calculate the maximum power output from the reactor assuming that it is

100 %

efficient.

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 6

The equation shows the radioactive decay of radon- $222$ .

${}_{86}^{222}{Rn} \to {}_{84}^{218}{Po} + {}_{2}^{4}α + γ$

Calculate the total energy output from this decay and state what forms of energy are produced.

(Mass of ${}_{86}^{222}{Rn} = 221.970 u$ , mass of ${}_{84}^{222}{Po} = 217.963 u$ , mass of ${}_{2}^{4}α = 4.002 u, 1 u$ is the unified atomic mass unit $= 1.660 \times 10^{- 27} kg .)$

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 7

A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit $(u)$ is defined as $\frac{1}{12}$ the mass of the carbon-12 atom. Calculate:

(a) the mass defect in kilograms

(Mass of a proton $= 1.007276 u$ , mass of a neutron $= 1.008665 u$ , mass of an electron $= 0.000548 u$ .)

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 7

A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit $(u)$ is defined as $\frac{1}{12}$ the mass of the carbon-12 atom. Calculate:

(b) the binding energy

(Mass of a proton $= 1.007276 u$ , mass of a neutron $= 1.008665 u$ , mass of an electron $= 0.000548 u$ .)

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 7

A carbon-12 atom consists of six protons, six neutrons and six electrons. The unified atomic mass unit $(u)$ is defined as $\frac{1}{12}$ the mass of the carbon-12 atom. Calculate:

(Mass of a proton $= 1.007276 u$ , mass of a neutron $= 1.008665 u$ , mass of an electron $= 0.000548 u$ .)

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 8

The fusion reaction that holds most promise for the generation of electricity is the fusion of tritium ${}_{1}^{3}H$ and deuterium ${}_{1}^{2}H$ .

The following equation shows the process: ${}_{1}^{3}H + {}_{1}^{2}H \to {}_{2}^{4}{He} + {}_{1}^{1}H$

Calculate:

(a) the change in mass in the reaction

(Mass of ${}_{1}^{3}H = 3.015500 u$ , Mass of ${}_{1}^{2}H = 2.013553 u$ , Mass of ${}_{2}^{4}{He} = 4.00150 u$ , Mass of ${}_{1}^{1}H = 1.007276 u .$ )

HARD

AS and A Level

IMPORTANT

Physics for Cambridge International AS & A Level Coursebook 3rd Edition Digital Access>Chapter 29 - Nuclear Physics>EXAM-STYLE QUESTIONS>Q 8

The fusion reaction that holds most promise for the generation of electricity is the fusion of tritium ${}_{1}^{3}H$ and deuterium ${}_{1}^{2}H$ .

The following equation shows the process: ${}_{1}^{3}H + {}_{1}^{2}H \to {}_{2}^{4}{He} + {}_{1}^{1}H$

Calculate:

(b) the energy released in the reaction

(Mass of ${}_{1}^{3}H = 3.015500 u$ , Mass of ${}_{1}^{2}H = 2.013553 u$ , Mass of ${}_{2}^{4}{He} = 4.00150 u$ , Mass of ${}_{1}^{1}H = 1.007276 u .$ )

Important Questions on Nuclear Physics

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