13.13 The radionuclide <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">C</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mmultiscripts></math> decays according to <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">C</mi><mi mathvariant="bold-italic"></mi><mo>→</mo><mi mathvariant="bold-italic"></mi><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">B</mi><mo>+</mo><mi mathvariant="bold-italic"></mi><msup><mrow><mrow><mi mathvariant="bold-italic">e</mi></mrow></mrow><mrow><mrow><mo>+</mo></mrow></mrow></msup><mo>+</mo><mi mathvariant="bold-italic"></mi><mi mathvariant="bold-italic">v</mi><mi mathvariant="bold-italic"></mi><mo>:</mo><mi mathvariant="bold-italic"></mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>5</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mmultiscripts></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mmultiscripts><msub><mrow><mrow><mi mathvariant="bold-italic">T</mi></mrow></mrow><mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></mrow></msub></math> =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">C</mi><mo>)</mo></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mmultiscripts></math> = 11.011434 u and m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">B</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>5</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mmultiscripts></math> ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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

13.13 The radionuclide ${}_{6}^{11}C$ decays according to ${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v :}} T_{1 / 2}$ =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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Payal Gupta | Contributor-Level 10

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13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

=&nb

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13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

= $[11.011434 - 11.009305 - 2 \times 0.000548] c^{2}$ u

=1.033 $\times 10^{- 3} c^{2}$ u

But 1 u = 931.5 MeV/ $c^{2}$

ΔQ = 0.962 MeV

The value of $Δ Q$ is almost comparable to the maximum energy of the emitted positron.

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13.13 The radionuclide C611 decays according to C→B+e++v:511611T1/2 =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( C)611 = 11.011434 u and m ( B511 ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.