QBManager

NUCLEAR PHYSICS

147463 For nuclei with mass number close to 119 and 238, the binding energies per nucleon are approximately 7.6 $MeV$ and $8.6 MeV$ , respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?

1

214 MeV

2

119 MeV

3

2047 MeV

4

1142 MeV

Explanation:

A For a nuclei mass number 119 and 238 binding energy per nucleon is $7.6 MeV & 8.6 MeV$
$_{92} U^{238} \to 2 (X^{119}) + Q$
$∴$ Binding energy of mass number $119 = 119 \times 7.6$ $= 904.4 MeV$
And, binding energy of $_{92} U^{238} = 238 \times 8.6 MeV$
$= 2046.8 MeV$
$∵$ Energy released $(Q) = (BE)_{Reactant} - (BE)_{Product}$
$Q = (2046.8 - 2 \times 904.4) MeV (_{92} U^{238} = 2 X^{119} + Q)$
$Q = 2046.8 - 1808.8$
$Q = 238 MeV$
$Q \approx 214 MeV$

WB JEE 2020

NUCLEAR PHYSICS

147464 Find $BE$ per nucleon of $^{56} Fe$ where $m (^{56} Fe) =$ $55.936 u, m_{n} = 1.008664 u, m_{p} = 1.007276 u$

1

477.45 MeV

2

8.52 MeV

3

577 MeV

4

10.52 MeV

Explanation:

B Given that,
$m_{n} = 1.008644 u, m_{p} = 1.007276 u$
$BE = [26 m_{p} + 30 m_{n} - m (^{56} Fe)] c^{2}$
$BE = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \times 931$
$= 0.51282 \times 931$
$= 477.435$
$= 477$
Binding energy per nucleon $= \frac{477.435}{56} MeV$
$= 8.52 MeV$

AIIMS-26.05.2019(M) Shift-1

NUCLEAR PHYSICS

147465 If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleons and that of $N^{15}$ is $7.7 MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

1

5.25 MeV

2

0.2 MeV

3

10.5 MeV

4

0.4 MeV

Explanation:

C Binding energy per nucleon of $N^{14} = 7.5 MeV$ Binding energy per nucleon of $N^{15} = 7.7 MeV$
Total binding energy of $N^{15} = 7.7 \times 15 MeV$
$= 115.5 MeV$
$Total binding energy of N^{14} = 7.5 \times 14 MeV$
$= 105 MeV$
Energy required to remove a neutron from $N^{15}$ is
$= 115.5 - 105$
$= 10.5 MeV$

AP EAMCET (21.04.2019) Shift-II

NUCLEAR PHYSICS

147466 The mass density of a nucleus varies with mass number $A$ as

1

A^{2}

2

A^{1}

3

A^{0}

4

A^{- 1}

Explanation:

C Atomic mass $=$ density $\times$ volume of atom
$R = R_{0} (A)^{\frac{1}{3}}$
We know that,
$ρ = \frac{Mass}{Volume}$
$ρ = \frac{A}{\frac{4}{3} π {(1.2 \times 10^{- 15}) \times (A)^{\frac{1}{3}}}^{3}}$
$ρ = \frac{4}{3} π {(1.2 \times 10^{- 15})}^{3} \times A^{0}$
$ρ = A^{0} \times constant$

UPSEE 2019

NUCLEAR PHYSICS

147468 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass numbers 80 and 120 . If binding energy per nucleon for parent atom is $6.5 MeV$ and for daughter nuclii is $7 MeV$ and $8 MeV$ respectively, then the energy released in the decay will be

1

200 MeV

2

120 MeV

3

220 MeV

4

180 Mev

Explanation:

C Given that
$X^{200} \to X^{80} + X^{120} + Q$
Binding per nucleon for parent atom $6.5 MeV$ and daughter nuclei $7 MeV$ and $8 MeV$
Energy released $(Q) = (BE)_{Product} - (BE)_{Reactant}$
$Q = 80 \times 7 + 120 \times 8 - 6.5 \times 200$
$Q = 220 MeV$

AP EAMCET-26.04.2017

NUCLEAR PHYSICS

147463 For nuclei with mass number close to 119 and 238, the binding energies per nucleon are approximately 7.6 $MeV$ and $8.6 MeV$ , respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?

1

214 MeV

2

119 MeV

3

2047 MeV

4

1142 MeV

Explanation:

A For a nuclei mass number 119 and 238 binding energy per nucleon is $7.6 MeV & 8.6 MeV$
$_{92} U^{238} \to 2 (X^{119}) + Q$
$∴$ Binding energy of mass number $119 = 119 \times 7.6$ $= 904.4 MeV$
And, binding energy of $_{92} U^{238} = 238 \times 8.6 MeV$
$= 2046.8 MeV$
$∵$ Energy released $(Q) = (BE)_{Reactant} - (BE)_{Product}$
$Q = (2046.8 - 2 \times 904.4) MeV (_{92} U^{238} = 2 X^{119} + Q)$
$Q = 2046.8 - 1808.8$
$Q = 238 MeV$
$Q \approx 214 MeV$

WB JEE 2020

NUCLEAR PHYSICS

147464 Find $BE$ per nucleon of $^{56} Fe$ where $m (^{56} Fe) =$ $55.936 u, m_{n} = 1.008664 u, m_{p} = 1.007276 u$

1

477.45 MeV

2

8.52 MeV

3

577 MeV

4

10.52 MeV

Explanation:

B Given that,
$m_{n} = 1.008644 u, m_{p} = 1.007276 u$
$BE = [26 m_{p} + 30 m_{n} - m (^{56} Fe)] c^{2}$
$BE = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \times 931$
$= 0.51282 \times 931$
$= 477.435$
$= 477$
Binding energy per nucleon $= \frac{477.435}{56} MeV$
$= 8.52 MeV$

AIIMS-26.05.2019(M) Shift-1

NUCLEAR PHYSICS

147465 If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleons and that of $N^{15}$ is $7.7 MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

1

5.25 MeV

2

0.2 MeV

3

10.5 MeV

4

0.4 MeV

Explanation:

C Binding energy per nucleon of $N^{14} = 7.5 MeV$ Binding energy per nucleon of $N^{15} = 7.7 MeV$
Total binding energy of $N^{15} = 7.7 \times 15 MeV$
$= 115.5 MeV$
$Total binding energy of N^{14} = 7.5 \times 14 MeV$
$= 105 MeV$
Energy required to remove a neutron from $N^{15}$ is
$= 115.5 - 105$
$= 10.5 MeV$

AP EAMCET (21.04.2019) Shift-II

NUCLEAR PHYSICS

147466 The mass density of a nucleus varies with mass number $A$ as

1

A^{2}

2

A^{1}

3

A^{0}

4

A^{- 1}

Explanation:

C Atomic mass $=$ density $\times$ volume of atom
$R = R_{0} (A)^{\frac{1}{3}}$
We know that,
$ρ = \frac{Mass}{Volume}$
$ρ = \frac{A}{\frac{4}{3} π {(1.2 \times 10^{- 15}) \times (A)^{\frac{1}{3}}}^{3}}$
$ρ = \frac{4}{3} π {(1.2 \times 10^{- 15})}^{3} \times A^{0}$
$ρ = A^{0} \times constant$

UPSEE 2019

NUCLEAR PHYSICS

147468 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass numbers 80 and 120 . If binding energy per nucleon for parent atom is $6.5 MeV$ and for daughter nuclii is $7 MeV$ and $8 MeV$ respectively, then the energy released in the decay will be

1

200 MeV

2

120 MeV

3

220 MeV

4

180 Mev

Explanation:

C Given that
$X^{200} \to X^{80} + X^{120} + Q$
Binding per nucleon for parent atom $6.5 MeV$ and daughter nuclei $7 MeV$ and $8 MeV$
Energy released $(Q) = (BE)_{Product} - (BE)_{Reactant}$
$Q = 80 \times 7 + 120 \times 8 - 6.5 \times 200$
$Q = 220 MeV$

AP EAMCET-26.04.2017

NUCLEAR PHYSICS

147463 For nuclei with mass number close to 119 and 238, the binding energies per nucleon are approximately 7.6 $MeV$ and $8.6 MeV$ , respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?

1

214 MeV

2

119 MeV

3

2047 MeV

4

1142 MeV

Explanation:

A For a nuclei mass number 119 and 238 binding energy per nucleon is $7.6 MeV & 8.6 MeV$
$_{92} U^{238} \to 2 (X^{119}) + Q$
$∴$ Binding energy of mass number $119 = 119 \times 7.6$ $= 904.4 MeV$
And, binding energy of $_{92} U^{238} = 238 \times 8.6 MeV$
$= 2046.8 MeV$
$∵$ Energy released $(Q) = (BE)_{Reactant} - (BE)_{Product}$
$Q = (2046.8 - 2 \times 904.4) MeV (_{92} U^{238} = 2 X^{119} + Q)$
$Q = 2046.8 - 1808.8$
$Q = 238 MeV$
$Q \approx 214 MeV$

WB JEE 2020

NUCLEAR PHYSICS

147464 Find $BE$ per nucleon of $^{56} Fe$ where $m (^{56} Fe) =$ $55.936 u, m_{n} = 1.008664 u, m_{p} = 1.007276 u$

1

477.45 MeV

2

8.52 MeV

3

577 MeV

4

10.52 MeV

Explanation:

B Given that,
$m_{n} = 1.008644 u, m_{p} = 1.007276 u$
$BE = [26 m_{p} + 30 m_{n} - m (^{56} Fe)] c^{2}$
$BE = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \times 931$
$= 0.51282 \times 931$
$= 477.435$
$= 477$
Binding energy per nucleon $= \frac{477.435}{56} MeV$
$= 8.52 MeV$

AIIMS-26.05.2019(M) Shift-1

NUCLEAR PHYSICS

147465 If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleons and that of $N^{15}$ is $7.7 MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

1

5.25 MeV

2

0.2 MeV

3

10.5 MeV

4

0.4 MeV

Explanation:

C Binding energy per nucleon of $N^{14} = 7.5 MeV$ Binding energy per nucleon of $N^{15} = 7.7 MeV$
Total binding energy of $N^{15} = 7.7 \times 15 MeV$
$= 115.5 MeV$
$Total binding energy of N^{14} = 7.5 \times 14 MeV$
$= 105 MeV$
Energy required to remove a neutron from $N^{15}$ is
$= 115.5 - 105$
$= 10.5 MeV$

AP EAMCET (21.04.2019) Shift-II

NUCLEAR PHYSICS

147466 The mass density of a nucleus varies with mass number $A$ as

1

A^{2}

2

A^{1}

3

A^{0}

4

A^{- 1}

Explanation:

C Atomic mass $=$ density $\times$ volume of atom
$R = R_{0} (A)^{\frac{1}{3}}$
We know that,
$ρ = \frac{Mass}{Volume}$
$ρ = \frac{A}{\frac{4}{3} π {(1.2 \times 10^{- 15}) \times (A)^{\frac{1}{3}}}^{3}}$
$ρ = \frac{4}{3} π {(1.2 \times 10^{- 15})}^{3} \times A^{0}$
$ρ = A^{0} \times constant$

UPSEE 2019

NUCLEAR PHYSICS

147468 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass numbers 80 and 120 . If binding energy per nucleon for parent atom is $6.5 MeV$ and for daughter nuclii is $7 MeV$ and $8 MeV$ respectively, then the energy released in the decay will be

1

200 MeV

2

120 MeV

3

220 MeV

4

180 Mev

Explanation:

C Given that
$X^{200} \to X^{80} + X^{120} + Q$
Binding per nucleon for parent atom $6.5 MeV$ and daughter nuclei $7 MeV$ and $8 MeV$
Energy released $(Q) = (BE)_{Product} - (BE)_{Reactant}$
$Q = 80 \times 7 + 120 \times 8 - 6.5 \times 200$
$Q = 220 MeV$

AP EAMCET-26.04.2017

NUCLEAR PHYSICS

147463 For nuclei with mass number close to 119 and 238, the binding energies per nucleon are approximately 7.6 $MeV$ and $8.6 MeV$ , respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?

1

214 MeV

2

119 MeV

3

2047 MeV

4

1142 MeV

Explanation:

A For a nuclei mass number 119 and 238 binding energy per nucleon is $7.6 MeV & 8.6 MeV$
$_{92} U^{238} \to 2 (X^{119}) + Q$
$∴$ Binding energy of mass number $119 = 119 \times 7.6$ $= 904.4 MeV$
And, binding energy of $_{92} U^{238} = 238 \times 8.6 MeV$
$= 2046.8 MeV$
$∵$ Energy released $(Q) = (BE)_{Reactant} - (BE)_{Product}$
$Q = (2046.8 - 2 \times 904.4) MeV (_{92} U^{238} = 2 X^{119} + Q)$
$Q = 2046.8 - 1808.8$
$Q = 238 MeV$
$Q \approx 214 MeV$

WB JEE 2020

NUCLEAR PHYSICS

147464 Find $BE$ per nucleon of $^{56} Fe$ where $m (^{56} Fe) =$ $55.936 u, m_{n} = 1.008664 u, m_{p} = 1.007276 u$

1

477.45 MeV

2

8.52 MeV

3

577 MeV

4

10.52 MeV

Explanation:

B Given that,
$m_{n} = 1.008644 u, m_{p} = 1.007276 u$
$BE = [26 m_{p} + 30 m_{n} - m (^{56} Fe)] c^{2}$
$BE = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \times 931$
$= 0.51282 \times 931$
$= 477.435$
$= 477$
Binding energy per nucleon $= \frac{477.435}{56} MeV$
$= 8.52 MeV$

AIIMS-26.05.2019(M) Shift-1

NUCLEAR PHYSICS

147465 If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleons and that of $N^{15}$ is $7.7 MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

1

5.25 MeV

2

0.2 MeV

3

10.5 MeV

4

0.4 MeV

Explanation:

C Binding energy per nucleon of $N^{14} = 7.5 MeV$ Binding energy per nucleon of $N^{15} = 7.7 MeV$
Total binding energy of $N^{15} = 7.7 \times 15 MeV$
$= 115.5 MeV$
$Total binding energy of N^{14} = 7.5 \times 14 MeV$
$= 105 MeV$
Energy required to remove a neutron from $N^{15}$ is
$= 115.5 - 105$
$= 10.5 MeV$

AP EAMCET (21.04.2019) Shift-II

NUCLEAR PHYSICS

147466 The mass density of a nucleus varies with mass number $A$ as

1

A^{2}

2

A^{1}

3

A^{0}

4

A^{- 1}

Explanation:

C Atomic mass $=$ density $\times$ volume of atom
$R = R_{0} (A)^{\frac{1}{3}}$
We know that,
$ρ = \frac{Mass}{Volume}$
$ρ = \frac{A}{\frac{4}{3} π {(1.2 \times 10^{- 15}) \times (A)^{\frac{1}{3}}}^{3}}$
$ρ = \frac{4}{3} π {(1.2 \times 10^{- 15})}^{3} \times A^{0}$
$ρ = A^{0} \times constant$

UPSEE 2019

NUCLEAR PHYSICS

147468 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass numbers 80 and 120 . If binding energy per nucleon for parent atom is $6.5 MeV$ and for daughter nuclii is $7 MeV$ and $8 MeV$ respectively, then the energy released in the decay will be

1

200 MeV

2

120 MeV

3

220 MeV

4

180 Mev

Explanation:

C Given that
$X^{200} \to X^{80} + X^{120} + Q$
Binding per nucleon for parent atom $6.5 MeV$ and daughter nuclei $7 MeV$ and $8 MeV$
Energy released $(Q) = (BE)_{Product} - (BE)_{Reactant}$
$Q = 80 \times 7 + 120 \times 8 - 6.5 \times 200$
$Q = 220 MeV$

AP EAMCET-26.04.2017

NUCLEAR PHYSICS

147463 For nuclei with mass number close to 119 and 238, the binding energies per nucleon are approximately 7.6 $MeV$ and $8.6 MeV$ , respectively. If a nucleus of mass number 238 breaks into two nuclei of nearly equal masses, what will be the approximate amount of energy released in the process of fission?

1

214 MeV

2

119 MeV

3

2047 MeV

4

1142 MeV

Explanation:

A For a nuclei mass number 119 and 238 binding energy per nucleon is $7.6 MeV & 8.6 MeV$
$_{92} U^{238} \to 2 (X^{119}) + Q$
$∴$ Binding energy of mass number $119 = 119 \times 7.6$ $= 904.4 MeV$
And, binding energy of $_{92} U^{238} = 238 \times 8.6 MeV$
$= 2046.8 MeV$
$∵$ Energy released $(Q) = (BE)_{Reactant} - (BE)_{Product}$
$Q = (2046.8 - 2 \times 904.4) MeV (_{92} U^{238} = 2 X^{119} + Q)$
$Q = 2046.8 - 1808.8$
$Q = 238 MeV$
$Q \approx 214 MeV$

WB JEE 2020

NUCLEAR PHYSICS

147464 Find $BE$ per nucleon of $^{56} Fe$ where $m (^{56} Fe) =$ $55.936 u, m_{n} = 1.008664 u, m_{p} = 1.007276 u$

1

477.45 MeV

2

8.52 MeV

3

577 MeV

4

10.52 MeV

Explanation:

B Given that,
$m_{n} = 1.008644 u, m_{p} = 1.007276 u$
$BE = [26 m_{p} + 30 m_{n} - m (^{56} Fe)] c^{2}$
$BE = [26 \times 1.00727 + 30 \times 1.00866 - 55.936] \times 931$
$= 0.51282 \times 931$
$= 477.435$
$= 477$
Binding energy per nucleon $= \frac{477.435}{56} MeV$
$= 8.52 MeV$

AIIMS-26.05.2019(M) Shift-1

NUCLEAR PHYSICS

147465 If the binding energy of $N^{14}$ is $7.5 MeV$ per nucleons and that of $N^{15}$ is $7.7 MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

1

5.25 MeV

2

0.2 MeV

3

10.5 MeV

4

0.4 MeV

Explanation:

C Binding energy per nucleon of $N^{14} = 7.5 MeV$ Binding energy per nucleon of $N^{15} = 7.7 MeV$
Total binding energy of $N^{15} = 7.7 \times 15 MeV$
$= 115.5 MeV$
$Total binding energy of N^{14} = 7.5 \times 14 MeV$
$= 105 MeV$
Energy required to remove a neutron from $N^{15}$ is
$= 115.5 - 105$
$= 10.5 MeV$

AP EAMCET (21.04.2019) Shift-II

NUCLEAR PHYSICS

147466 The mass density of a nucleus varies with mass number $A$ as

1

A^{2}

2

A^{1}

3

A^{0}

4

A^{- 1}

Explanation:

C Atomic mass $=$ density $\times$ volume of atom
$R = R_{0} (A)^{\frac{1}{3}}$
We know that,
$ρ = \frac{Mass}{Volume}$
$ρ = \frac{A}{\frac{4}{3} π {(1.2 \times 10^{- 15}) \times (A)^{\frac{1}{3}}}^{3}}$
$ρ = \frac{4}{3} π {(1.2 \times 10^{- 15})}^{3} \times A^{0}$
$ρ = A^{0} \times constant$

UPSEE 2019

NUCLEAR PHYSICS

147468 A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass numbers 80 and 120 . If binding energy per nucleon for parent atom is $6.5 MeV$ and for daughter nuclii is $7 MeV$ and $8 MeV$ respectively, then the energy released in the decay will be

1

200 MeV

2

120 MeV

3

220 MeV

4

180 Mev

Explanation:

C Given that
$X^{200} \to X^{80} + X^{120} + Q$
Binding per nucleon for parent atom $6.5 MeV$ and daughter nuclei $7 MeV$ and $8 MeV$
Energy released $(Q) = (BE)_{Product} - (BE)_{Reactant}$
$Q = 80 \times 7 + 120 \times 8 - 6.5 \times 200$
$Q = 220 MeV$

AP EAMCET-26.04.2017