QBManager

PHXII13:NUCLEI

363772 A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 $u$ . The energy liberated per $u$ is
(given $1 u = 931 M e V$ )

1

2.67 M e V

2

26.7 M e V

3

6.675 M e V

4

13.35 M e V

Explanation:

Energy liberated $E = Δ m c^{2}$
$= 0.028664 u \times 931 \frac{M e V}{u} = 26.7 M e V$
As four nucleons are involved in the reaction the energy liberated per each nucleon (1 $u$ ) is
$\frac{26.7 M e V}{4} = 6.675 M e V$

PHXII13:NUCLEI

363773 The ratio of the amounts of energy released as a result of the fusion of $1 k g$ hydrogen $(E_{1})$ and fission of $1 k g$ of $_{92} U^{236} (E_{2})$ will be (Given energy released fusion of Hydrogen is $28 M e V$ and that released per fission of $U^{236}$ is $200 M e V$ )

1 4.13

2 3.28

3 5.28

4 1.28

Explanation:

For Hydrogen, 4 atoms involve in one fusion
$\Rightarrow$ number of fusions in $1 k g$ of Hydrogen $= N_{1}$
$\Rightarrow N_{1} = \frac{500 \times N}{4} = 125 N$
Similarly $N_{2} \to$ no. of fissions involved in $1 k g$ of $U^{236}$ .
$\Rightarrow N_{2} = \frac{1000}{236} \times N$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{125 \times N \times 28}{\frac{1000}{236} \times N \times 200}$
$= \frac{125 \times 28 \times 236}{1000 \times 200}$
$= 4.13$

PHXII13:NUCLEI

363774 Inside the sun

1 Four nuclei of hydrogen combine to form two nuclei of helium

2 Four nuclei of hydrogen combine to form four nuclei of helium

3 Four nuclei of hydrogen combine to form one nucleus of helium

4 Four nuclei of hydrogen is transformed into one nucleus of helium

PHXII13:NUCLEI

363775 An atomic power nuclear reactor can deliver $300 M W$ . The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 M e V$ . The number of uranium atoms fissioned per hour will be

1

4 \times 10^{22}

2

30 \times 1 0^{25}

3

5 \times 1 0^{15}

4

10 \times 1 0^{20}

Explanation:

$P = \frac{N E}{t} \Rightarrow 300 \times 10^{6}$
$= \frac{N \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$
Number of atoms per hour $= 1.02 \times 10^{19} \times 3600$
$= 3.97 \times 10^{22} .$

PHXII13:NUCLEI

363776 In a nuclear reactor of the function of the moderator is to decrease

1 Number of neutrons

2 Speed of neutrons

3 Escape of neutrons

4 Temperature of the reactor

KCET - 2017

PHXII13:NUCLEI

363772 A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 $u$ . The energy liberated per $u$ is
(given $1 u = 931 M e V$ )

1

2.67 M e V

2

26.7 M e V

3

6.675 M e V

4

13.35 M e V

Explanation:

Energy liberated $E = Δ m c^{2}$
$= 0.028664 u \times 931 \frac{M e V}{u} = 26.7 M e V$
As four nucleons are involved in the reaction the energy liberated per each nucleon (1 $u$ ) is
$\frac{26.7 M e V}{4} = 6.675 M e V$

PHXII13:NUCLEI

363773 The ratio of the amounts of energy released as a result of the fusion of $1 k g$ hydrogen $(E_{1})$ and fission of $1 k g$ of $_{92} U^{236} (E_{2})$ will be (Given energy released fusion of Hydrogen is $28 M e V$ and that released per fission of $U^{236}$ is $200 M e V$ )

1 4.13

2 3.28

3 5.28

4 1.28

Explanation:

For Hydrogen, 4 atoms involve in one fusion
$\Rightarrow$ number of fusions in $1 k g$ of Hydrogen $= N_{1}$
$\Rightarrow N_{1} = \frac{500 \times N}{4} = 125 N$
Similarly $N_{2} \to$ no. of fissions involved in $1 k g$ of $U^{236}$ .
$\Rightarrow N_{2} = \frac{1000}{236} \times N$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{125 \times N \times 28}{\frac{1000}{236} \times N \times 200}$
$= \frac{125 \times 28 \times 236}{1000 \times 200}$
$= 4.13$

PHXII13:NUCLEI

363774 Inside the sun

1 Four nuclei of hydrogen combine to form two nuclei of helium

2 Four nuclei of hydrogen combine to form four nuclei of helium

3 Four nuclei of hydrogen combine to form one nucleus of helium

4 Four nuclei of hydrogen is transformed into one nucleus of helium

PHXII13:NUCLEI

363775 An atomic power nuclear reactor can deliver $300 M W$ . The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 M e V$ . The number of uranium atoms fissioned per hour will be

1

4 \times 10^{22}

2

30 \times 1 0^{25}

3

5 \times 1 0^{15}

4

10 \times 1 0^{20}

Explanation:

$P = \frac{N E}{t} \Rightarrow 300 \times 10^{6}$
$= \frac{N \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$
Number of atoms per hour $= 1.02 \times 10^{19} \times 3600$
$= 3.97 \times 10^{22} .$

PHXII13:NUCLEI

363776 In a nuclear reactor of the function of the moderator is to decrease

1 Number of neutrons

2 Speed of neutrons

3 Escape of neutrons

4 Temperature of the reactor

KCET - 2017

PHXII13:NUCLEI

363772 A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 $u$ . The energy liberated per $u$ is
(given $1 u = 931 M e V$ )

1

2.67 M e V

2

26.7 M e V

3

6.675 M e V

4

13.35 M e V

Explanation:

Energy liberated $E = Δ m c^{2}$
$= 0.028664 u \times 931 \frac{M e V}{u} = 26.7 M e V$
As four nucleons are involved in the reaction the energy liberated per each nucleon (1 $u$ ) is
$\frac{26.7 M e V}{4} = 6.675 M e V$

PHXII13:NUCLEI

363773 The ratio of the amounts of energy released as a result of the fusion of $1 k g$ hydrogen $(E_{1})$ and fission of $1 k g$ of $_{92} U^{236} (E_{2})$ will be (Given energy released fusion of Hydrogen is $28 M e V$ and that released per fission of $U^{236}$ is $200 M e V$ )

1 4.13

2 3.28

3 5.28

4 1.28

Explanation:

For Hydrogen, 4 atoms involve in one fusion
$\Rightarrow$ number of fusions in $1 k g$ of Hydrogen $= N_{1}$
$\Rightarrow N_{1} = \frac{500 \times N}{4} = 125 N$
Similarly $N_{2} \to$ no. of fissions involved in $1 k g$ of $U^{236}$ .
$\Rightarrow N_{2} = \frac{1000}{236} \times N$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{125 \times N \times 28}{\frac{1000}{236} \times N \times 200}$
$= \frac{125 \times 28 \times 236}{1000 \times 200}$
$= 4.13$

PHXII13:NUCLEI

363774 Inside the sun

1 Four nuclei of hydrogen combine to form two nuclei of helium

2 Four nuclei of hydrogen combine to form four nuclei of helium

3 Four nuclei of hydrogen combine to form one nucleus of helium

4 Four nuclei of hydrogen is transformed into one nucleus of helium

PHXII13:NUCLEI

363775 An atomic power nuclear reactor can deliver $300 M W$ . The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 M e V$ . The number of uranium atoms fissioned per hour will be

1

4 \times 10^{22}

2

30 \times 1 0^{25}

3

5 \times 1 0^{15}

4

10 \times 1 0^{20}

Explanation:

$P = \frac{N E}{t} \Rightarrow 300 \times 10^{6}$
$= \frac{N \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$
Number of atoms per hour $= 1.02 \times 10^{19} \times 3600$
$= 3.97 \times 10^{22} .$

PHXII13:NUCLEI

363776 In a nuclear reactor of the function of the moderator is to decrease

1 Number of neutrons

2 Speed of neutrons

3 Escape of neutrons

4 Temperature of the reactor

KCET - 2017

PHXII13:NUCLEI

363772 A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 $u$ . The energy liberated per $u$ is
(given $1 u = 931 M e V$ )

1

2.67 M e V

2

26.7 M e V

3

6.675 M e V

4

13.35 M e V

Explanation:

Energy liberated $E = Δ m c^{2}$
$= 0.028664 u \times 931 \frac{M e V}{u} = 26.7 M e V$
As four nucleons are involved in the reaction the energy liberated per each nucleon (1 $u$ ) is
$\frac{26.7 M e V}{4} = 6.675 M e V$

PHXII13:NUCLEI

363773 The ratio of the amounts of energy released as a result of the fusion of $1 k g$ hydrogen $(E_{1})$ and fission of $1 k g$ of $_{92} U^{236} (E_{2})$ will be (Given energy released fusion of Hydrogen is $28 M e V$ and that released per fission of $U^{236}$ is $200 M e V$ )

1 4.13

2 3.28

3 5.28

4 1.28

Explanation:

For Hydrogen, 4 atoms involve in one fusion
$\Rightarrow$ number of fusions in $1 k g$ of Hydrogen $= N_{1}$
$\Rightarrow N_{1} = \frac{500 \times N}{4} = 125 N$
Similarly $N_{2} \to$ no. of fissions involved in $1 k g$ of $U^{236}$ .
$\Rightarrow N_{2} = \frac{1000}{236} \times N$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{125 \times N \times 28}{\frac{1000}{236} \times N \times 200}$
$= \frac{125 \times 28 \times 236}{1000 \times 200}$
$= 4.13$

PHXII13:NUCLEI

363774 Inside the sun

1 Four nuclei of hydrogen combine to form two nuclei of helium

2 Four nuclei of hydrogen combine to form four nuclei of helium

3 Four nuclei of hydrogen combine to form one nucleus of helium

4 Four nuclei of hydrogen is transformed into one nucleus of helium

PHXII13:NUCLEI

363775 An atomic power nuclear reactor can deliver $300 M W$ . The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 M e V$ . The number of uranium atoms fissioned per hour will be

1

4 \times 10^{22}

2

30 \times 1 0^{25}

3

5 \times 1 0^{15}

4

10 \times 1 0^{20}

Explanation:

$P = \frac{N E}{t} \Rightarrow 300 \times 10^{6}$
$= \frac{N \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$
Number of atoms per hour $= 1.02 \times 10^{19} \times 3600$
$= 3.97 \times 10^{22} .$

PHXII13:NUCLEI

363776 In a nuclear reactor of the function of the moderator is to decrease

1 Number of neutrons

2 Speed of neutrons

3 Escape of neutrons

4 Temperature of the reactor

KCET - 2017

PHXII13:NUCLEI

363772 A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is 0.02866 $u$ . The energy liberated per $u$ is
(given $1 u = 931 M e V$ )

1

2.67 M e V

2

26.7 M e V

3

6.675 M e V

4

13.35 M e V

Explanation:

Energy liberated $E = Δ m c^{2}$
$= 0.028664 u \times 931 \frac{M e V}{u} = 26.7 M e V$
As four nucleons are involved in the reaction the energy liberated per each nucleon (1 $u$ ) is
$\frac{26.7 M e V}{4} = 6.675 M e V$

PHXII13:NUCLEI

363773 The ratio of the amounts of energy released as a result of the fusion of $1 k g$ hydrogen $(E_{1})$ and fission of $1 k g$ of $_{92} U^{236} (E_{2})$ will be (Given energy released fusion of Hydrogen is $28 M e V$ and that released per fission of $U^{236}$ is $200 M e V$ )

1 4.13

2 3.28

3 5.28

4 1.28

Explanation:

For Hydrogen, 4 atoms involve in one fusion
$\Rightarrow$ number of fusions in $1 k g$ of Hydrogen $= N_{1}$
$\Rightarrow N_{1} = \frac{500 \times N}{4} = 125 N$
Similarly $N_{2} \to$ no. of fissions involved in $1 k g$ of $U^{236}$ .
$\Rightarrow N_{2} = \frac{1000}{236} \times N$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{125 \times N \times 28}{\frac{1000}{236} \times N \times 200}$
$= \frac{125 \times 28 \times 236}{1000 \times 200}$
$= 4.13$

PHXII13:NUCLEI

363774 Inside the sun

1 Four nuclei of hydrogen combine to form two nuclei of helium

2 Four nuclei of hydrogen combine to form four nuclei of helium

3 Four nuclei of hydrogen combine to form one nucleus of helium

4 Four nuclei of hydrogen is transformed into one nucleus of helium

PHXII13:NUCLEI

363775 An atomic power nuclear reactor can deliver $300 M W$ . The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 M e V$ . The number of uranium atoms fissioned per hour will be

1

4 \times 10^{22}

2

30 \times 1 0^{25}

3

5 \times 1 0^{15}

4

10 \times 1 0^{20}

Explanation:

$P = \frac{N E}{t} \Rightarrow 300 \times 10^{6}$
$= \frac{N \times 170 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$∴$ Number of atoms per sec $\frac{N}{t} = 1.102 \times 10^{19}$
Number of atoms per hour $= 1.02 \times 10^{19} \times 3600$
$= 3.97 \times 10^{22} .$

PHXII13:NUCLEI

363776 In a nuclear reactor of the function of the moderator is to decrease

1 Number of neutrons

2 Speed of neutrons

3 Escape of neutrons

4 Temperature of the reactor

KCET - 2017