13.12 Find the Q-value and the kinetic energy of the emitted <math><mi mathvariant="bold-italic">α</mi></math> -particle in the <math><mi mathvariant="bold-italic">α</mi></math> -decay of (a) <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">R</mi><mi mathvariant="bold-italic">a</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>88</mn></mrow></mrow><mrow><mrow><mn>226</mn></mrow></mrow></mmultiscripts></math> and (b) <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">R</mi><mi mathvariant="bold-italic">n</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>86</mn></mrow></mrow><mrow><mrow><mn>220</mn></mrow></mrow></mmultiscripts></math> . Given m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">R</mi><mi mathvariant="bold-italic">a</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>88</mn></mrow></mrow><mrow><mrow><mn>226</mn></mrow></mrow></mmultiscripts><mo>)</mo></math> = 226.02540 u, m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">R</mi><mi mathvariant="bold-italic">n</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>86</mn></mrow></mrow><mrow><mrow><mn>222</mn></mrow></mrow></mmultiscripts><mo>)</mo></math> = 222.01750 u, m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">R</mi><mi mathvariant="bold-italic">n</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>86</mn></mrow></mrow><mrow><mrow><mn>222</mn></mrow></mrow></mmultiscripts><mo>)</mo><mi mathvariant="bold-italic"></mi></math> = 220.01137 u, m ( <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">P</mi><mi mathvariant="bold-italic">o</mi><mo>)</mo><mi mathvariant="bold-italic"></mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>84</mn></mrow></mrow><mrow><mrow><mn>216</mn></mrow></mrow></mmultiscripts></math> = 216.00189 u.

13.12α- particle decay of Ra88226 emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.Ra88226 → Rn86222 + He24Q value of emitted α- particle = (Sum of initial mass – Sum of final mass) ×c2 , wherec = Speed of light.It is given thatm ( Ra88226) = 226.02540 um ( Rn86222) = 222.01750 um ( He24) = 4.002603 uQ value = [(226.02540) – (222.01750 + 4.002603)] c2= 5.297 ×10-3uc2But 1 u = 931.5 MeV/ c2Hence Q = 4.934 MeVKinetic energy of the α- particle = MassnumberafterdecayMassnumberbeforedecay ×Q = 222226 × 4.934= 4.85 MeVα- particle decay of Rn86220 emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.Rn86220 → Po84216 + He24Q value of emitted α- particle = (Sum of initial mass – Sum of final mass) ×c2 , wherec = Speed of light.It is given thatm ( Rn86222) = 220.01137 um ( Po)84216 = 216.00189 um ( He24) = 4.002603 uQ value = [(220.01137) – (216.00189 + 4.002603)] c2= 6.877 ×10-3uc2But 1 u = 931.5 MeV/ c2Hence Q = 6.406 MeVKinetic energy of the α- particle = MassnumberafterdecayMassnumberbeforedecay ×Q = 216222 × 6.406= 6.23 MeV

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

13.12 Find the Q-value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of (a) ${}_{88}^{226}{R a}$ and (b) ${}_{86}^{220}{R n}$ . Given m ( ${}_{88}^{226}{R a})$ = 226.02540 u, m ( ${}_{86}^{222}{R n})$ = 222.01750 u, m ( ${}_{86}^{222}{R n})$ = 220.01137 u, m ( ${}_{84}^{216}{P o)}$ = 216.00189 u.

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13.12 $α -$ particle decay of ${}_{88}^{226}{R a}$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $\times c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{88}^{226}{R a})$ = 226.02540 u

m ( ${}_{86}^{222}{R n})$ = 222.01750 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^{2}$

= 5.297 $\times 10^{- 3} {u c}^{2}$

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

Hence Q = 4.934 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $\times Q$ = $\frac{222}{226}$ $\times$ 4.934= 4.85 MeV

$α -$ particle decay

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13.12 $α -$ particle decay of ${}_{88}^{226}{R a}$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $\times c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{88}^{226}{R a})$ = 226.02540 u

m ( ${}_{86}^{222}{R n})$ = 222.01750 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^{2}$

= 5.297 $\times 10^{- 3} {u c}^{2}$

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

Hence Q = 4.934 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $\times Q$ = $\frac{222}{226}$ $\times$ 4.934= 4.85 MeV

$α -$ particle decay of ${}_{86}^{220}{R n}$ emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.

${}_{86}^{220}{R n}$ $\to$ ${}_{84}^{216}{P o}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $\times c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{86}^{222}{R n})$ = 220.01137 u

m ( ${}_{84}^{216}{P o)}$ = 216.00189 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(220.01137) – (216.00189 + 4.002603)] $c^{2}$

= 6.877 $\times 10^{- 3} {u c}^{2}$

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

Hence Q = 6.406 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $\times Q$ = $\frac{216}{222}$ $\times$ 6.406= 6.23 MeV

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13.12 Find the Q-value and the kinetic energy of the emitted α -particle in the α -decay of (a) Ra88226 and (b) Rn86220 . Given m ( Ra88226) = 226.02540 u, m ( Rn86222) = 222.01750 u, m ( Rn86222) = 220.01137 u, m ( Po)84216 = 216.00189 u.

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