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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}^{235}U$ in a fission reactor.

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Payal Gupta | Contributor-Level 10

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13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_{1}^{1}{H)}$ contains 6.023 $* 10^{23}$ atoms

1 kg of hydrogen contains = 1000 $*$ 6.023 $* 10^{23}$ atoms = 6.023 $* 10^{26}$ atoms

Within Sun, four ( ${}_{1}^{1}{H)}$ nuclei combine and forms 1 ${}_{2}^{4}{H e}$ nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_{1}^{1}H$ is:

$E_{1}$ = $\frac{6.023 * 10^{26} * 26}{4}$ MeV = 3.91495 $* 10^{27}$ MeV

Amount of ${}_{92}^{235}U$ , m = 1 kg = 1000 g

1 mole of ${}_{92}^{235}U$ , i.e. 235 g of ${}_{92}^{235}U$ contains 6.023 $* 10^{23}$ atoms

1 kg of ${}_{92}^{235}U$ contains = $\frac{1000}{235}$ $*$ 6.023 $* 10^{23}$ atoms = 2.563 $* 10^{24}$ atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}^{235}U$ is 200

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13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_{1}^{1}{H)}$ contains 6.023 $* 10^{23}$ atoms

1 kg of hydrogen contains = 1000 $*$ 6.023 $* 10^{23}$ atoms = 6.023 $* 10^{26}$ atoms

Within Sun, four ( ${}_{1}^{1}{H)}$ nuclei combine and forms 1 ${}_{2}^{4}{H e}$ nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_{1}^{1}H$ is:

$E_{1}$ = $\frac{6.023 * 10^{26} * 26}{4}$ MeV = 3.91495 $* 10^{27}$ MeV

Amount of ${}_{92}^{235}U$ , m = 1 kg = 1000 g

1 mole of ${}_{92}^{235}U$ , i.e. 235 g of ${}_{92}^{235}U$ contains 6.023 $* 10^{23}$ atoms

1 kg of ${}_{92}^{235}U$ contains = $\frac{1000}{235}$ $*$ 6.023 $* 10^{23}$ atoms = 2.563 $* 10^{24}$ atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}^{235}U$ is 200 MeV.

Hence the energy released from fusion of 1 kg of ${}_{92}^{235}U$ is:

= 2.563 $* 10^{24} * 200$ MeV = 5.126 $* 10^{26}$ MeV

$\frac{E_{1}}{E_{2}}$ = $\frac{3.91495 * 10^{27}}{5.126 * 10^{26}}$ = 7.64

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of ${}_{92}^{235}U$ .

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13.30 Amount of hydrogen, m = 1 kg = 1000 g1 mole of hydrogen, i.e. 1 g of hydrogen ( <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math> contains 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math>  atoms1 kg of hydrogen contains = 1000 <math> <mo>×</mo> </math> 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math>  atoms = 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </math> atomsWithin Sun, four ( <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math>  nuclei combine and forms 1 <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mi>e</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus. In this process 26 MeV of energy is released.Hence the energy released from fusion of 1 kg of <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math>  is: <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math>  = <math> <mfrac> <mrow> <mrow> <mn>6.023</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>26</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math>  MeV = 3.91495 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </msup> </math> MeVAmount of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math> , m = 1 kg = 1000 g1 mole of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math> , i.e. 235 g of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  contains 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> atoms1 kg of  <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math> contains = <math> <mfrac> <mrow> <mrow> <mn>1000</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mfrac> </math> <math> <mo>×</mo> </math>  6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> atoms = 2.563 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> </math>  atomsIt is known that the amount of energy released in the fission of 1 atom of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  is 200 MeV.Hence the energy released from fusion of 1 kg of  <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math>  is: = 2.563 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>200</mn> </math>  MeV = 5.126 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </math> MeV <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>3.91495</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>5.126</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 7.64Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of <math> <mmultiscripts> <mrow> <mrow> <mi>U</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>92</mn> </mrow> </mrow> <mrow> <mrow> <mn>235</mn> </mrow> </mrow> </mmultiscripts> </math> .

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13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}^{235}U$ in a fission reactor.

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13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of U 92 235 in a fission reactor.

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13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}^{235}U$ in a fission reactor.