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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

(a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_{1}^{2}{H)}$ =2.014102 u

m( ${}_{1}^{3}{H)}$ =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

(Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzmann’s constant, T = absolute temperature.)

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Payal Gupta | Contributor-Level 10

3 months ago

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $\times 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $\times 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 \times 10^{- 15}}$ $\times$ 9 $\times 10^{9}$ = 5.76 $\times 10^{- 14}$ J =

...more

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $\times 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $\times 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 \times 10^{- 15}}$ $\times$ 9 $\times 10^{9}$ = 5.76 $\times 10^{- 14}$ J = $\frac{5.76 \times 10^{- 14}}{1.6 \times 10^{- 19}}$ eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $\times \frac{3}{2} k T$

Where

k = Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{K E}{3 k}$ = $\frac{5.76 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}}$ = 1.39 $\times 10^{9}$ K

Hence the gas must be heated to 1.39 $\times 10^{9}$ K to initiate the reaction.

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13.28 The equation for deuterium-tritium fusion is given as: <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mo>+</mo> <mi> </mi> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mi> </mi> <mo>→</mo> <mi> </mi> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mi>e</mi> <mo>+</mo> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mmultiscripts> </math>  It is given thatMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi mathvariant="normal">H</mi> <mo>)</mo> <mo>,</mo> <mi mathvariant="normal"> </mi> <msub> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mmultiscripts> </math>  = 2.014102 uMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi mathvariant="normal">H</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <msub> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>=</mo> <mi mathvariant="normal"> </mi> </math> 3.016049 uMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mi>e</mi> <mo>)</mo> <mo>,</mo> <mi> </mi> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mmultiscripts> </math>  = 4.002603 uMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> <mo>)</mo> <mo>,</mo> <mi> </mi> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math> = 1.008665 uQ-value of the given D-T reaction is:Q = <math> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mn>2.014102</mn> <mo>+</mo> <mn>3</mn> <mo>.</mo> <mn>016049</mn> <mo>-</mo> <mn>4.002603</mn> <mo>-</mo> <mn>1.008665</mn> </mrow> </mrow> </mfenced> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>u</mi> </math> = 0.018883 <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>u</mi> </math> = 17.59 MeVRadius of the deuterium and tritium, r <math> <mo>≈</mo> <mn>2.0</mn> <mi> </mi> <mi>f</mi> <mi>m</mi> </math> = 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </math> mDistance between the centers of the nucleus when they touch each other,d = r +r = 4 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </math> mCharge on the deuterium and tritium nucleus = eHence the repulsive potential energy between the two nuclei is given as:V = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi>π</mi> <msub> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>d</mi> </mrow> </mrow> </mfrac> </math> Where, <math> <msub> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </math>  = permittivity of free space It is given that <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mi>π</mi> <msub> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math>  = 9 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> </math> N <math> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> Hence, V = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> <math> <mo>×</mo> </math>  9 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> </math> = 5.76 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </math>  J = <math> <mfrac> <mrow> <mrow> <mn>5.76</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  eV = 360 keVHence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.It is given thatKE = 2 <math> <mo>×</mo> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>k</mi> <mi>T</mi> </math> Wherek = Boltzmann constant = 1.38 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> </math> <math> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>k</mi> <mi>g</mi> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> T = Temperature required to trigger the reactionTherefore T = <math> <mfrac> <mrow> <mrow> <mi>K</mi> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <mi>k</mi> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>5.76</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> <mo>×</mo> <mn>1.38</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 1.39 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> </math>  KHence the gas must be heated to 1.39 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> </math> K to initiate the reaction.

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