The binding energy per nucleon of O 16 8 is 7 . 97 M e V and that of O 17 8 is 7 . 75 M e V . Find the energy needed to remove a neutron from O 17 8 .

Given:The binding energy per nucleon of O 16 8 is 7 . 97 MeV , the binding energy will be 16 × 7 . 97 = 127 . 52 MeV and given binding energy per nucleon O 17 8 is 7 . 75 MeV , the binding energy will be 17 × 7 . 75 MeV = 131 . 75 MeV The binding energy of the nucleus is given as: BE = [ Z m H + ( A - Z ) m n - m ( X A Z ) ] × C 2 , where the letters have their usual meaning. The energy needed to remove a neutron from O 17 8 will be difference in the binding energy of O 16 8 and O 17 8 since the latter have one extra neutron in it. The required value will be 131 . 75 - 127 . 52 MeV = 4 . 23 MeV .

How much energy is needed to remove a neutron from the nucleus of A 13 l 27 ? Given: mass of A 13 l 27 atom = 26 . 981541 u , mass of A 13 l 26 atom = 25 . 986895 u , mass of neutron = 1 . 008665 u .

Given:The binding energy per nucleon of 13 A l 26 is 25 . 986895 MeV , the binding energy will be 26 × 25 . 986895 = 675 . 65927 MeV and given binding energy per nucleon 13 A l 27 is 7 . 75 MeV , the binding energy will be 17 × 7 . 75 MeV = 131 . 75 MeV The binding energy of the nucleus is given as: BE = [ Z m H + ( A - Z ) m n - m ( X A Z ) ] × C 2 , where the letters have their usual meaning. The energy needed to remove a neutron from Al 27 13 will be difference in the binding energy of A 13 l 26 and Al 27 13 since the latter have one extra neutron in it. The required value will be 131 . 75 - 127 . 52 MeV = 4 . 23 MeV .

100 g of L 3 i 7 is converted into H 2 e 4 by proton bombardment. Calculate the energy released. Given: mass of L 3 i 7 atom = 7 . 016 u , mass of H 2 e 4 atom = 4 . 004 u , mass of H 1 1 atom = 1 . 008 u . The Avogadro number is 6 . 02 × 10 23 mol - 1 .

Mass of L i 7 3 = 7 . 016 Mass of H 2 e 4 = 4 . 004 u Mass of H 1 1 = 1 . 008 Avogadro numbe r N A = 6 . 02 × 10 23 / mole The mass defect is given by , ∆ m = m ( Li ) + m ( p ) - 2 m ( He ) = 7 . 016 + 1 . 008 - 2 × 4 . 004 = 0 . 0184 u Number of atoms in 100 gm lithium are N = N A A × 100 = 6 . 023 × 10 23 7 × 100 = 8 . 6 × 10 24 The energy released Q = N × △ m × 931 MeV Or Q = 8 . 6 × 10 24 × 0 . 0188 × 931 Q = 8 . 6 × 10 24 × 0 . 0188 × 931 × 1 . 66 × 10 - 19 × 10 6 J = 2 × 10 13 J

E M B I B E

ISC Physics Class 12, Part-2>Chapter 16 - Mass-Energy Equivalence : Nuclear Binding Energy>QUESTIONS>Q 13

HARD

12th ICSE

IMPORTANT

Earn 100

If we form a lithium nucleus $({}_{3}L i^{7})$ by $4$ neutrons and $3$ protons, then how much energy will be released? $(m_{p} = 1.007825 u, m_{n} = 1.008665 u)$ the mass of the lithium nucleus is $7.01436 u$ .

Important Questions on Mass-Energy Equivalence : Nuclear Binding Energy

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 16 - Mass-Energy Equivalence : Nuclear Binding Energy>QUESTIONS>Q 14

The binding energy per nucleon of

{}_{8}O^{16}

is

7.97 M e V

and that of

{}_{8}O^{17}

is

7.75 M e V

. Find the energy needed to remove a neutron from

{}_{8}O^{17}

.

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 16 - Mass-Energy Equivalence : Nuclear Binding Energy>QUESTIONS>Q 15

Calculate the disintegration energy

Q

when a heavy nucleus of mass number

A = 240

with binding energy per nucleon

7.6 M e V

is split into two equal fragments of mass number

A' = 120

, each with

B E

/nucleon

8.5 M e V

.

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 16 - Mass-Energy Equivalence : Nuclear Binding Energy>QUESTIONS>Q 16

How much energy is needed to remove a neutron from the nucleus of

{}_{13}A l^{27}

? Given: mass of

{}_{13}A l^{27}

atom

= 26.981541 u

, mass of

{}_{13}A l^{26}

atom

= 25.986895 u

, mass of neutron

= 1.008665 u

.

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 16 - Mass-Energy Equivalence : Nuclear Binding Energy>QUESTIONS>Q 17

Bombardment of lithium with protons give rise to the following reaction:

${}_{3}L i^{7} + \underset{(proton)}{{}_{+ 1}H^{1}} \to \underset{(α - particle)}{{}_{2}H e^{4}} + {}_{2}H e^{4} + energy$ .

The atomic masses of lithium, hydrogen and helium are $7.016 u, 1.008 u$ and $4.004 u$ respectively. Find the energy carried by each $α$ -particle. Given: $1 u = 931 M e V$ .