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

13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ${}_{20}^{41}{C a}$ and from the following data:

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

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13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m(

...more

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

less

13.24 If a neutron <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> </math>  is removed from <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>41</mn> </mrow> </mrow> </mmultiscripts> </math> , the corresponding reaction can be written as: <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>41</mn> </mrow> </mrow> </mmultiscripts> <mi> </mi> <mo>→</mo> </math> <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>40</mn> </mrow> </mrow> </mmultiscripts> </math> + <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math> The separation energies areFor <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>41</mn> </mrow> </mrow> </mmultiscripts> </math>  : Separation energy = 8.363007 MeVFor <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </mmultiscripts> </math>  : Separation energy = 13.059 MeVIt is given thatm( <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>40</mn> </mrow> </mrow> </mmultiscripts> </math>  = 39.962591 um( <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>41</mn> </mrow> </mrow> </mmultiscripts> </math> = 40.962278 um( <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> </math> = 1.008665 u The mass defect of the reaction is given as:Δm = m <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>40</mn> </mrow> </mrow> </mmultiscripts> <mo>+</mo> </math> (  m( <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> <mo>-</mo> </math> m( <math> <mmultiscripts> <mrow> <mrow> <mi>C</mi> <mi>a</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>41</mn> </mrow> </mrow> </mmultiscripts> </math> = 39.962591 + 1.008665 - 40.962278= 8.978 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> u=8.363007 MeV For <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </mmultiscripts> </math> , the neutron removal reaction can be written as <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </mmultiscripts> <mo>→</mo> </math> <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </mmultiscripts> </math> + <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> </math> It is given thatm( <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </mmultiscripts> </math> = 25.986895 um( <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </mmultiscripts> </math>  = 26.981541 um( <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> </math> = 1.008665 uThe mass defect of the reaction is given as:Δm = m( <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>26</mn> </mrow> </mrow> </mmultiscripts> <mo>+</mo> </math>  m( <math> <mmultiscripts> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> <mo>-</mo> </math> m( <math> <mmultiscripts> <mrow> <mrow> <mi>A</mi> <mi>l</mi> <mo>)</mo> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>13</mn> </mrow> </mrow> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> </mmultiscripts> </math> = 25.986895 + 1.008665 - 26.981541= 0.014019 u=13.0586985 MeV

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