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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.29 Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of decays in the decay scheme shown in Fig. 13.6. You are given that

m( ${}^{198}{A u}$ ) = 197.968233 u

m( ${}^{198}{H g}$ ) =197.966760 u

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Payal Gupta | Contributor-Level 10

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13.29 It can be observed from the given $γ -$ decay diagram that $γ_{1}$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to decay is given as:

= 1.088 – 0 = 1.088 MeV = 1.088 $\times 10^{6}$ eV = 1.088 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 1.7408 $\times 10^{- 13}$ J

We know, $E_{1} = h ν_{1}$ , where

$ν_{1}$ = Frequency of radiation radiated by $γ_{1}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{1}$ = $\frac{E_{1}}{h}$ = $\frac{1.7408 \times 10^{- 13}}{6.6 \times 10^{- 34}}$ = 2.637 $\times 10^{20}$ Hz

It can be observed from the given $γ -$ decay diagram that $γ_{2}$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to $γ_{2}$ decay is given as:

$E_{2}$ = 0.412 – 0 = 0.412 MeV = 0.412 $\times 10^{6}$ eV = 0.412 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 6.

...more

13.29 It can be observed from the given $γ -$ decay diagram that $γ_{1}$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to decay is given as:

= 1.088 – 0 = 1.088 MeV = 1.088 $\times 10^{6}$ eV = 1.088 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 1.7408 $\times 10^{- 13}$ J

We know, $E_{1} = h ν_{1}$ , where

$ν_{1}$ = Frequency of radiation radiated by $γ_{1}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{1}$ = $\frac{E_{1}}{h}$ = $\frac{1.7408 \times 10^{- 13}}{6.6 \times 10^{- 34}}$ = 2.637 $\times 10^{20}$ Hz

It can be observed from the given $γ -$ decay diagram that $γ_{2}$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to $γ_{2}$ decay is given as:

$E_{2}$ = 0.412 – 0 = 0.412 MeV = 0.412 $\times 10^{6}$ eV = 0.412 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 6.592 $\times 10^{- 14}$ J

We know, $E_{2} = h ν_{2}$ , where

$ν_{2}$ = Frequency of radiation radiated by $γ_{2}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{2}$ = $\frac{E_{2}}{h}$ = $\frac{6.592 \times 10^{- 14}}{6.6 \times 10^{- 34}}$ = 9.987 $\times 10^{19}$ Hz

It can be observed from the given $γ -$ decay diagram that $γ_{3}$ decays from 1.008 MeV energy level to the 0.412 MeV energy level.

Hence the energy corresponding to decay is given as:

= 1.088 - 0.412 = 0.676 MeV = 0.676 $\times 10^{6}$ eV = 0.676 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 1.0816 $\times 10^{- 13}$ J

We know, $E_{3} = h ν_{3}$ , where

$ν_{3}$ = Frequency of radiation radiated by $γ_{3}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{3}$ = $\frac{E_{3}}{h}$ = $\frac{1.0816 \times 10^{- 13}}{6.6 \times 10^{- 34}}$ = 1.639 $\times 10^{20}$ Hz

Mass of ( ${}_{79}^{198}{A u}$ ) = 197.968233 u

Mass of ( ${}_{79}^{198}{A u}$ ) =197.966760 u

Energy of the highest level is given as:

E = $[m ({}_{79}^{198}{A u}) - m ({}_{80}^{198}{H g})] c^{2}$

= $[197.968233 - 197.966760] c^{2}$ u

= 1.473 $\times 10^{- 3} c^{2}$ u

= 1.473 $\times 10^{- 3} \times 931.5$ MeV

= 1.3720995 MeV

$β_{1}^{-} d e c a y s$ from 1.3720995 MeV level to 1.088 MeV level

Hence maximum kinetic energy of the $β_{1}^{-}$ particles = 1.3720995 - 1.088 = 0.2840995 MeV

$β_{2}^{-}$ from 1.3720995 MeV level to 0.412 MeV level

Hence maximum kinetic energy of the $β_{2}^{-}$ particles = 1.3720995 – 0.412 = 0.9600995 MeV

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13.29 It can be observed from the given <math> <mi>γ</mi> <mo>-</mo> </math> decay diagram that <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> decays from 1.088 MeV energy level to the 0 MeV energy level.Hence the energy corresponding to  decay is given as: = 1.088 – 0 = 1.088 MeV = 1.088 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math>  eV = 1.088 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> J= 1.7408 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>13</mn> </mrow> </mrow> </msup> </math> JWe know, <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>=</mo> <mi>h</mi> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> , where <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = Frequency of radiation radiated by <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math>  decay <math> <mi>h</mi> <mo>=</mo> <mi mathvariant="normal">P</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <msup> <mrow> <mrow> <mi mathvariant="normal">k</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">t</mi> <mi> </mi> </math> = 6.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math> JsHence,  <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>1.7408</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>13</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>6.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 2.637 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msup> </math> HzIt can be observed from the given <math> <mi>γ</mi> <mo>-</mo> </math> decay diagram that <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> decays from 0.412 MeV energy level to the 0 MeV energy level.Hence the energy corresponding to <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  decay is given as: <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = 0.412 – 0 = 0.412 MeV = 0.412 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math>  eV = 0.412 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math>  J= 6.592 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </math> JWe know, <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>=</mo> <mi>h</mi> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> , where <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> = Frequency of radiation radiated by <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> decay <math> <mi>h</mi> <mo>=</mo> <mi mathvariant="normal">P</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <msup> <mrow> <mrow> <mi mathvariant="normal">k</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">t</mi> <mi> </mi> </math> = 6.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math> JsHence, <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math>  = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>6.592</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>6.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 9.987 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>19</mn> </mrow> </mrow> </msup> </math>  HzIt can be observed from the given <math> <mi>γ</mi> <mo>-</mo> </math> decay diagram that <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> decays from 1.008 MeV energy level to the 0.412 MeV energy level.Hence the energy corresponding to  decay is given as: = 1.088 - 0.412 = 0.676 MeV = 0.676 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math>  eV = 0.676 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math>  J= 1.0816 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>13</mn> </mrow> </mrow> </msup> </math>  JWe know, <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> <mo>=</mo> <mi>h</mi> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> , where <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math> = Frequency of radiation radiated by <math> <msub> <mrow> <mrow> <mi>γ</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math>  decay <math> <mi>h</mi> <mo>=</mo> <mi mathvariant="normal">P</mi> <mi mathvariant="normal">l</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <msup> <mrow> <mrow> <mi mathvariant="normal">k</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">t</mi> <mi> </mi> </math> = 6.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math>  Js Hence, <math> <msub> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </math>  = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>1.0816</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>13</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>6.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 1.639 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msup> </math> HzMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi mathvariant="normal">A</mi> <mi mathvariant="normal">u</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>79</mn> </mrow> </mrow> <mrow> <mrow> <mn>198</mn> </mrow> </mrow> </mmultiscripts> </math> ) = 197.968233 uMass of ( <math> <mmultiscripts> <mrow> <mrow> <mi mathvariant="normal">A</mi> <mi mathvariant="normal">u</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>79</mn> </mrow> </mrow> <mrow> <mrow> <mn>198</mn> </mrow> </mrow> </mmultiscripts> </math> ) =197.966760 uEnergy of the highest level is given as:E = <math> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mi mathvariant="normal">m</mi> <mfenced separators="|"> <mrow> <mrow> <mmultiscripts> <mrow> <mrow> <mi mathvariant="normal">A</mi> <mi mathvariant="normal">u</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>79</mn> </mrow> </mrow> <mrow> <mrow> <mn>198</mn> </mrow> </mrow> </mmultiscripts> </mrow> </mrow> </mfenced> <mo>-</mo> <mi mathvariant="normal">m</mi> <mo>(</mo> <mmultiscripts> <mrow> <mrow> <mi>H</mi> <mi>g</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>80</mn> </mrow> </mrow> <mrow> <mrow> <mn>198</mn> </mrow> </mrow> </mmultiscripts> <mo>)</mo> <mi mathvariant="normal"> </mi> </mrow> </mrow> </mfenced> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mn>197.968233</mn> <mo>-</mo> <mn>197.966760</mn> </mrow> </mrow> </mfenced> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> u= 1.473 <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= 1.473 <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= 1.3720995 MeV <math> <msubsup> <mrow> <mrow> <mi>β</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> </mrow> </mrow> </msubsup> <mi> </mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">y</mi> <mi mathvariant="normal">s</mi> </math>  from 1.3720995 MeV level to 1.088 MeV levelHence maximum kinetic energy of the <math> <msubsup> <mrow> <mrow> <mi>β</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> </mrow> </mrow> </msubsup> </math> particles = 1.3720995 - 1.088 = 0.2840995 MeV <math> <msubsup> <mrow> <mrow> <mi>β</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> </mrow> </mrow> </msubsup> </math> from 1.3720995 MeV level to 0.412 MeV levelHence maximum kinetic energy of the <math> <msubsup> <mrow> <mrow> <mi>β</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> </mrow> </mrow> </msubsup> </math> particles = 1.3720995 – 0.412 = 0.9600995 MeV

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13.29 Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of decays in the decay scheme shown in Fig. 13.6. You are given that

m( ${}^{198}{A u}$ ) = 197.968233 u

m( ${}^{198}{H g}$ ) =197.966760 u

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13.29 Obtain the maximum kinetic energy of β -particles, and the radiation frequencies of decays in the decay scheme shown in Fig. 13.6. You are given that m( A u 198 ) = 197.968233 u m( H g 198 ) =197.966760 u

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13.29 Obtain the maximum kinetic energy of $β$ -particles, and the radiation frequencies of decays in the decay scheme shown in Fig. 13.6. You are given that

m( ${}^{198}{A u}$ ) = 197.968233 u

m( ${}^{198}{H g}$ ) =197.966760 u