Ask & Answer Question

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.

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

3 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.

less

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.

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question