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

13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $α -$ particle. Consider the following decay processes:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

Calculate the Q-values for these decays and determine that both are energetically allowed.

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Payal Gupta | Contributor-Level 10

5 months ago

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ &n

...more

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 219.00948 - 4.00260) $c^{2}$ u

= 6.42 $\times 10^{- 3} c^{2}$ u

= 6.42 $\times 10^{- 3} \times 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

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13.26 For the emission of <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> <mrow> <mrow> <mn>14</mn> </mrow> </mrow> </mmultiscripts> </math> , the nuclear reaction is: <math> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>a</mi> <mi> </mi> <mo>→</mo> <mmultiscripts> <mrow> <mrow> <mi>P</mi> <mi>b</mi> <mo>+</mo> <mi> </mi> <mmultiscripts> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> <mrow> <mrow> <mn>14</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>82</mn> </mrow> </mrow> <mrow> <mrow> <mn>209</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>88</mn> </mrow> </mrow> <mrow> <mrow> <mn>223</mn> </mrow> </mrow> </mmultiscripts> </math> We know that:Mass of <math> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>88</mn> </mrow> </mrow> <mrow> <mrow> <mn>223</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 223.01850 uMass of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> <mi>b</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>82</mn> </mrow> </mrow> <mrow> <mrow> <mn>209</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  = 208.98107 uMass of <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> <mrow> <mrow> <mn>14</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <mi> </mi> <mi> </mi> <mi> </mi> <mi> </mi> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> = 14.00324 uHence, the Q-value of the reaction is given as:Q = ( <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> - <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  - <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> ) <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = (223.01850 - 208.98107 - 14.00324) <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> u= 0.03419 <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> u= 0.03419 MeV = 31.848 MeVHence, the Q-value of the nuclear reaction is 31.848 MeV, since the value is positive, the reaction is energetically allowed.For the emission of <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> , the nuclear reaction is: <math> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>a</mi> <mi> </mi> <mo>→</mo> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>n</mi> <mo>+</mo> <mi> </mi> <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> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>86</mn> </mrow> </mrow> <mrow> <mrow> <mn>219</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>88</mn> </mrow> </mrow> <mrow> <mrow> <mn>223</mn> </mrow> </mrow> </mmultiscripts> </math> We know that:Mass of <math> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>88</mn> </mrow> </mrow> <mrow> <mrow> <mn>223</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = 223.01850 uMass of <math> <mmultiscripts> <mrow> <mrow> <mi>R</mi> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>86</mn> </mrow> </mrow> <mrow> <mrow> <mn>219</mn> </mrow> </mrow> </mmultiscripts> </math> , <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  = 219.00948 uMass of <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> , <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> = 4.00260 uHence, the Q-value of the reaction is given as:Q = ( <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> - <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  - <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> ) <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = (223.01850 - 219.00948 - 4.00260) <math> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> u= 6.42 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> u= 6.42 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>931.5</mn> </math> MeV = 5.980 MeVHence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

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