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CBSE Class 12 Physics 2017 Delhi Set 1 Paper

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Question : 19 of 26
Marks: +1, -0
(i) A radioactive nucleus ' AA ' undergoes a series of decays as given below:
AαA1βA2αA3γA4A \overset{\alpha}{\rightarrow} A_1 \xrightarrow[]{β}A_2 \xrightarrow[]{α}A_3 \xrightarrow[]{γ}A_4
The mass number and atomic number of A2A_2 are 176 and 71 respectively.
Determine the mass and atomic numbers of A4A_4 and AA.
(ii) Write the basic nuclear processes underlying β+β^{+}and ββ^{-}decays.
Solution:  
(i) Determining the mass and atomic number of A4A_4 and AA
(ii) Basic nuclear processes of β+β^{+}and ββ^{-}decays
(i) A4:A_4: Mass Number : 172
Atomic Number : 69
A : Mass Number : 180
Atomic Number : 72
[Alternatively : Give full credit if student considers decay and find atomic and mass numbers accordingly
  72180Aα70176A1β+  71176A2α\; {}_{72}^{180} A \xrightarrow[]{ \alpha } {}_{70}^{176} A_1 \xrightarrow[]{ \beta^+ } \; {}_{71}^{176} A_2 \xrightarrow[]{ \alpha }   69172A3γ  69172A4\; {}_{69}^{172} A_3 \xrightarrow[]{ \gamma } \; {}_{69}^{172} A_4
Gives the values quoted above.
If the student takes β+β^{+}decay
  72180Aα70176A1β+  71176A2α\; {}_{72}^{180} A \xrightarrow[]{\alpha} {}_{70}^{176} A_1 \xrightarrow[]{\beta^{+}} \; {}_{71}^{176} A_2 \xrightarrow[]{\alpha}   69172A3γ  69172A4\; {}_{69}^{172} A_3 \xrightarrow[]{\gamma} \; {}_{69}^{172} A_4
This would give the answers: (A4:172,69);(A:180,74)]( A_4: 172,69) ;( A: 180,74)]
(ii) Basic nuclear process for β+β^{+}decay pn+10ep \rightarrow n + {}_{1}^{0} e +v+v
For ββ^{-}decay np+10e+vn \rightarrow p + {}_{-1}^{0} e + \overline{v}
[Note: Give full credit of this part, if student writes the processes as conversion of proton into neutron for decay and neutron into proton for decay.]
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