QBManager

NUCLEAR PHYSICS

148017 If the binding energy per nuclear in ${Li}^{7}$ and ${He}^{4}$ nuclei are respectively $5.60 Me V$ and 7.06 $MeV$ , then energy of reactor
${L i}^{7} + P \to 2_{2} {H e}^{4} is$

1

19.6 MeV

2

2.4 MeV

3

8.4 MeV

4

17.3 MeV

Explanation:

D Given that,
Binding energy of $({Li}^{7}) = 7 \times 5.60 MeV = 39.20 MeV$
And, binding energy of ${He}^{4} = 4 \times 7.06 MeV = 28.24$ $MeV$
On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times {He}^{4} -$ energy of ${Li}^{7}$
$= 2 \times 28.24 - 39.20$
$= 56.48 - 39.20$
$= 17.28 MeV$
$= 17.3 MeV$

VITEEE-2012

NUCLEAR PHYSICS

148022 On bombardment of $U^{235}$ by slow neutrons, 200 $MeV$ energy is released. If the power output of atomic reactor is $1.6 MW$ , then the rate of fission will be

1

5 \times 10^{16}

per second

2

10 \times 10^{16}

per second

3

15 \times 10^{16}

per second

4

20 \times 10^{16}

per second

Explanation:

A Given that
Energy release per fission $(E) = 200 MeV$
Output power of the reactor $(P) = 1.6 MW$
$∴ E = 200 \times 1.6 \times 10^{- 19} MJ$
$∵$ We know that
$Rate of fission (R) = \frac{P}{E}$
$∴ R = \frac{1.6}{200 \times 1.6 \times 10^{- 19}}$
$R = 5 \times 10^{16} per second$

J and K CET- 2005

NUCLEAR PHYSICS

148018 Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following

1 Fusion of two nuclei of mass number lying in the range of

100 < A < 200

will release energy

2 Fusion of two nuclei of mass number lying in the range of

51 < A < 100

will release energy

3 Fusion of two nuclei of mass number lying in the range of

1 < A < 50

will release energy

4 Fission of the nucleus of mass number lying in the range of

100 < A < 200

will release energy when broken into two fragments

Explanation:

B For fusion of two atoms of atomic mass number in the range $51 < A < 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.

Karnataka CET-2010

NUCLEAR PHYSICS

148024 In the fission of uranium, $0.5 g$ mass is decayed, then how much energy will be obtained by it?

1 1.25 kilo watt hour

2

1.25 \times 10^{7}

kilo watt hour

3 0.25 kilo watt hour

4

1.25 \times 10^{4}

kilo watt hour

Explanation:

B Given that,
Mass of decayed $(Δ m) = 0.5 g = 0.5 \times 10^{- 3} kg$
Energy released $(E) = Δ {mc}^{2}$
$E = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 4.5 \times 10^{13} J$
$E == \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$
$E = 1.25 \times 10^{7} kWh$

J and K CET- 2002

NUCLEAR PHYSICS

148026 In the nuclear fission reaction
$_{2}^{4} He +_{7}^{14} N \to_{p}^{q} X +_{1}^{1} H$
The nucleus $_{p}^{q} X$ is

1 nitrogen of mass 16

2 nitrogen of mass 17

3 oxygen of mass 16

4 oxygen of mass 17

Explanation:

D The given nuclear fission reaction,
$_{2}^{4} He +_{7}^{14} N \to_{p}^{9} X +_{1}^{1} H$
In order that reaction holds true mass number and atomic number remains conserved.
Mass number -
$4 + 14 = q + 1$
$q = 18 - 1$
$q = 17$
Atomic number -
$2 + 7 = p + 1$
$p = 9 - 1$
$p = 8$
Hence, unknown element is an isotope of oxygen of mass 17.

J and K CET- 2001

NUCLEAR PHYSICS

148017 If the binding energy per nuclear in ${Li}^{7}$ and ${He}^{4}$ nuclei are respectively $5.60 Me V$ and 7.06 $MeV$ , then energy of reactor
${L i}^{7} + P \to 2_{2} {H e}^{4} is$

1

19.6 MeV

2

2.4 MeV

3

8.4 MeV

4

17.3 MeV

Explanation:

D Given that,
Binding energy of $({Li}^{7}) = 7 \times 5.60 MeV = 39.20 MeV$
And, binding energy of ${He}^{4} = 4 \times 7.06 MeV = 28.24$ $MeV$
On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times {He}^{4} -$ energy of ${Li}^{7}$
$= 2 \times 28.24 - 39.20$
$= 56.48 - 39.20$
$= 17.28 MeV$
$= 17.3 MeV$

VITEEE-2012

NUCLEAR PHYSICS

148022 On bombardment of $U^{235}$ by slow neutrons, 200 $MeV$ energy is released. If the power output of atomic reactor is $1.6 MW$ , then the rate of fission will be

1

5 \times 10^{16}

per second

2

10 \times 10^{16}

per second

3

15 \times 10^{16}

per second

4

20 \times 10^{16}

per second

Explanation:

A Given that
Energy release per fission $(E) = 200 MeV$
Output power of the reactor $(P) = 1.6 MW$
$∴ E = 200 \times 1.6 \times 10^{- 19} MJ$
$∵$ We know that
$Rate of fission (R) = \frac{P}{E}$
$∴ R = \frac{1.6}{200 \times 1.6 \times 10^{- 19}}$
$R = 5 \times 10^{16} per second$

J and K CET- 2005

NUCLEAR PHYSICS

148018 Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following

1 Fusion of two nuclei of mass number lying in the range of

100 < A < 200

will release energy

2 Fusion of two nuclei of mass number lying in the range of

51 < A < 100

will release energy

3 Fusion of two nuclei of mass number lying in the range of

1 < A < 50

will release energy

4 Fission of the nucleus of mass number lying in the range of

100 < A < 200

will release energy when broken into two fragments

Explanation:

B For fusion of two atoms of atomic mass number in the range $51 < A < 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.

Karnataka CET-2010

NUCLEAR PHYSICS

148024 In the fission of uranium, $0.5 g$ mass is decayed, then how much energy will be obtained by it?

1 1.25 kilo watt hour

2

1.25 \times 10^{7}

kilo watt hour

3 0.25 kilo watt hour

4

1.25 \times 10^{4}

kilo watt hour

Explanation:

B Given that,
Mass of decayed $(Δ m) = 0.5 g = 0.5 \times 10^{- 3} kg$
Energy released $(E) = Δ {mc}^{2}$
$E = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 4.5 \times 10^{13} J$
$E == \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$
$E = 1.25 \times 10^{7} kWh$

J and K CET- 2002

NUCLEAR PHYSICS

148026 In the nuclear fission reaction
$_{2}^{4} He +_{7}^{14} N \to_{p}^{q} X +_{1}^{1} H$
The nucleus $_{p}^{q} X$ is

1 nitrogen of mass 16

2 nitrogen of mass 17

3 oxygen of mass 16

4 oxygen of mass 17

Explanation:

D The given nuclear fission reaction,
$_{2}^{4} He +_{7}^{14} N \to_{p}^{9} X +_{1}^{1} H$
In order that reaction holds true mass number and atomic number remains conserved.
Mass number -
$4 + 14 = q + 1$
$q = 18 - 1$
$q = 17$
Atomic number -
$2 + 7 = p + 1$
$p = 9 - 1$
$p = 8$
Hence, unknown element is an isotope of oxygen of mass 17.

J and K CET- 2001

NUCLEAR PHYSICS

148017 If the binding energy per nuclear in ${Li}^{7}$ and ${He}^{4}$ nuclei are respectively $5.60 Me V$ and 7.06 $MeV$ , then energy of reactor
${L i}^{7} + P \to 2_{2} {H e}^{4} is$

1

19.6 MeV

2

2.4 MeV

3

8.4 MeV

4

17.3 MeV

Explanation:

D Given that,
Binding energy of $({Li}^{7}) = 7 \times 5.60 MeV = 39.20 MeV$
And, binding energy of ${He}^{4} = 4 \times 7.06 MeV = 28.24$ $MeV$
On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times {He}^{4} -$ energy of ${Li}^{7}$
$= 2 \times 28.24 - 39.20$
$= 56.48 - 39.20$
$= 17.28 MeV$
$= 17.3 MeV$

VITEEE-2012

NUCLEAR PHYSICS

148022 On bombardment of $U^{235}$ by slow neutrons, 200 $MeV$ energy is released. If the power output of atomic reactor is $1.6 MW$ , then the rate of fission will be

1

5 \times 10^{16}

per second

2

10 \times 10^{16}

per second

3

15 \times 10^{16}

per second

4

20 \times 10^{16}

per second

Explanation:

A Given that
Energy release per fission $(E) = 200 MeV$
Output power of the reactor $(P) = 1.6 MW$
$∴ E = 200 \times 1.6 \times 10^{- 19} MJ$
$∵$ We know that
$Rate of fission (R) = \frac{P}{E}$
$∴ R = \frac{1.6}{200 \times 1.6 \times 10^{- 19}}$
$R = 5 \times 10^{16} per second$

J and K CET- 2005

NUCLEAR PHYSICS

148018 Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following

1 Fusion of two nuclei of mass number lying in the range of

100 < A < 200

will release energy

2 Fusion of two nuclei of mass number lying in the range of

51 < A < 100

will release energy

3 Fusion of two nuclei of mass number lying in the range of

1 < A < 50

will release energy

4 Fission of the nucleus of mass number lying in the range of

100 < A < 200

will release energy when broken into two fragments

Explanation:

B For fusion of two atoms of atomic mass number in the range $51 < A < 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.

Karnataka CET-2010

NUCLEAR PHYSICS

148024 In the fission of uranium, $0.5 g$ mass is decayed, then how much energy will be obtained by it?

1 1.25 kilo watt hour

2

1.25 \times 10^{7}

kilo watt hour

3 0.25 kilo watt hour

4

1.25 \times 10^{4}

kilo watt hour

Explanation:

B Given that,
Mass of decayed $(Δ m) = 0.5 g = 0.5 \times 10^{- 3} kg$
Energy released $(E) = Δ {mc}^{2}$
$E = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 4.5 \times 10^{13} J$
$E == \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$
$E = 1.25 \times 10^{7} kWh$

J and K CET- 2002

NUCLEAR PHYSICS

148026 In the nuclear fission reaction
$_{2}^{4} He +_{7}^{14} N \to_{p}^{q} X +_{1}^{1} H$
The nucleus $_{p}^{q} X$ is

1 nitrogen of mass 16

2 nitrogen of mass 17

3 oxygen of mass 16

4 oxygen of mass 17

Explanation:

D The given nuclear fission reaction,
$_{2}^{4} He +_{7}^{14} N \to_{p}^{9} X +_{1}^{1} H$
In order that reaction holds true mass number and atomic number remains conserved.
Mass number -
$4 + 14 = q + 1$
$q = 18 - 1$
$q = 17$
Atomic number -
$2 + 7 = p + 1$
$p = 9 - 1$
$p = 8$
Hence, unknown element is an isotope of oxygen of mass 17.

J and K CET- 2001

NUCLEAR PHYSICS

148017 If the binding energy per nuclear in ${Li}^{7}$ and ${He}^{4}$ nuclei are respectively $5.60 Me V$ and 7.06 $MeV$ , then energy of reactor
${L i}^{7} + P \to 2_{2} {H e}^{4} is$

1

19.6 MeV

2

2.4 MeV

3

8.4 MeV

4

17.3 MeV

Explanation:

D Given that,
Binding energy of $({Li}^{7}) = 7 \times 5.60 MeV = 39.20 MeV$
And, binding energy of ${He}^{4} = 4 \times 7.06 MeV = 28.24$ $MeV$
On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times {He}^{4} -$ energy of ${Li}^{7}$
$= 2 \times 28.24 - 39.20$
$= 56.48 - 39.20$
$= 17.28 MeV$
$= 17.3 MeV$

VITEEE-2012

NUCLEAR PHYSICS

148022 On bombardment of $U^{235}$ by slow neutrons, 200 $MeV$ energy is released. If the power output of atomic reactor is $1.6 MW$ , then the rate of fission will be

1

5 \times 10^{16}

per second

2

10 \times 10^{16}

per second

3

15 \times 10^{16}

per second

4

20 \times 10^{16}

per second

Explanation:

A Given that
Energy release per fission $(E) = 200 MeV$
Output power of the reactor $(P) = 1.6 MW$
$∴ E = 200 \times 1.6 \times 10^{- 19} MJ$
$∵$ We know that
$Rate of fission (R) = \frac{P}{E}$
$∴ R = \frac{1.6}{200 \times 1.6 \times 10^{- 19}}$
$R = 5 \times 10^{16} per second$

J and K CET- 2005

NUCLEAR PHYSICS

148018 Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following

1 Fusion of two nuclei of mass number lying in the range of

100 < A < 200

will release energy

2 Fusion of two nuclei of mass number lying in the range of

51 < A < 100

will release energy

3 Fusion of two nuclei of mass number lying in the range of

1 < A < 50

will release energy

4 Fission of the nucleus of mass number lying in the range of

100 < A < 200

will release energy when broken into two fragments

Explanation:

B For fusion of two atoms of atomic mass number in the range $51 < A < 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.

Karnataka CET-2010

NUCLEAR PHYSICS

148024 In the fission of uranium, $0.5 g$ mass is decayed, then how much energy will be obtained by it?

1 1.25 kilo watt hour

2

1.25 \times 10^{7}

kilo watt hour

3 0.25 kilo watt hour

4

1.25 \times 10^{4}

kilo watt hour

Explanation:

B Given that,
Mass of decayed $(Δ m) = 0.5 g = 0.5 \times 10^{- 3} kg$
Energy released $(E) = Δ {mc}^{2}$
$E = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 4.5 \times 10^{13} J$
$E == \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$
$E = 1.25 \times 10^{7} kWh$

J and K CET- 2002

NUCLEAR PHYSICS

148026 In the nuclear fission reaction
$_{2}^{4} He +_{7}^{14} N \to_{p}^{q} X +_{1}^{1} H$
The nucleus $_{p}^{q} X$ is

1 nitrogen of mass 16

2 nitrogen of mass 17

3 oxygen of mass 16

4 oxygen of mass 17

Explanation:

D The given nuclear fission reaction,
$_{2}^{4} He +_{7}^{14} N \to_{p}^{9} X +_{1}^{1} H$
In order that reaction holds true mass number and atomic number remains conserved.
Mass number -
$4 + 14 = q + 1$
$q = 18 - 1$
$q = 17$
Atomic number -
$2 + 7 = p + 1$
$p = 9 - 1$
$p = 8$
Hence, unknown element is an isotope of oxygen of mass 17.

J and K CET- 2001

NUCLEAR PHYSICS

148017 If the binding energy per nuclear in ${Li}^{7}$ and ${He}^{4}$ nuclei are respectively $5.60 Me V$ and 7.06 $MeV$ , then energy of reactor
${L i}^{7} + P \to 2_{2} {H e}^{4} is$

1

19.6 MeV

2

2.4 MeV

3

8.4 MeV

4

17.3 MeV

Explanation:

D Given that,
Binding energy of $({Li}^{7}) = 7 \times 5.60 MeV = 39.20 MeV$
And, binding energy of ${He}^{4} = 4 \times 7.06 MeV = 28.24$ $MeV$
On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times {He}^{4} -$ energy of ${Li}^{7}$
$= 2 \times 28.24 - 39.20$
$= 56.48 - 39.20$
$= 17.28 MeV$
$= 17.3 MeV$

VITEEE-2012

NUCLEAR PHYSICS

148022 On bombardment of $U^{235}$ by slow neutrons, 200 $MeV$ energy is released. If the power output of atomic reactor is $1.6 MW$ , then the rate of fission will be

1

5 \times 10^{16}

per second

2

10 \times 10^{16}

per second

3

15 \times 10^{16}

per second

4

20 \times 10^{16}

per second

Explanation:

A Given that
Energy release per fission $(E) = 200 MeV$
Output power of the reactor $(P) = 1.6 MW$
$∴ E = 200 \times 1.6 \times 10^{- 19} MJ$
$∵$ We know that
$Rate of fission (R) = \frac{P}{E}$
$∴ R = \frac{1.6}{200 \times 1.6 \times 10^{- 19}}$
$R = 5 \times 10^{16} per second$

J and K CET- 2005

NUCLEAR PHYSICS

148018 Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following

1 Fusion of two nuclei of mass number lying in the range of

100 < A < 200

will release energy

2 Fusion of two nuclei of mass number lying in the range of

51 < A < 100

will release energy

3 Fusion of two nuclei of mass number lying in the range of

1 < A < 50

will release energy

4 Fission of the nucleus of mass number lying in the range of

100 < A < 200

will release energy when broken into two fragments

Explanation:

B For fusion of two atoms of atomic mass number in the range $51 < A < 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.

Karnataka CET-2010

NUCLEAR PHYSICS

148024 In the fission of uranium, $0.5 g$ mass is decayed, then how much energy will be obtained by it?

1 1.25 kilo watt hour

2

1.25 \times 10^{7}

kilo watt hour

3 0.25 kilo watt hour

4

1.25 \times 10^{4}

kilo watt hour

Explanation:

B Given that,
Mass of decayed $(Δ m) = 0.5 g = 0.5 \times 10^{- 3} kg$
Energy released $(E) = Δ {mc}^{2}$
$E = 0.5 \times 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 4.5 \times 10^{13} J$
$E == \frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$
$E = 1.25 \times 10^{7} kWh$

J and K CET- 2002

NUCLEAR PHYSICS

148026 In the nuclear fission reaction
$_{2}^{4} He +_{7}^{14} N \to_{p}^{q} X +_{1}^{1} H$
The nucleus $_{p}^{q} X$ is

1 nitrogen of mass 16

2 nitrogen of mass 17

3 oxygen of mass 16

4 oxygen of mass 17

Explanation:

D The given nuclear fission reaction,
$_{2}^{4} He +_{7}^{14} N \to_{p}^{9} X +_{1}^{1} H$
In order that reaction holds true mass number and atomic number remains conserved.
Mass number -
$4 + 14 = q + 1$
$q = 18 - 1$
$q = 17$
Atomic number -
$2 + 7 = p + 1$
$p = 9 - 1$
$p = 8$
Hence, unknown element is an isotope of oxygen of mass 17.

J and K CET- 2001

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