Isobar

\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept, if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process  
\(p\rightarrow n+{ e }^{ + }+\upsilon \)

(a) \(\frac { dN }{ dt } =-\lambda N\)
\(\int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } =\int _{ 0 }^{ t }{ -\lambda dt } } \) 
\(\left[ { log }_{ e }^{ N } \right] _{ { N }_{ 0 } }^{ N }=-\lambda \left[ t \right] _{ 0 }^{ t }\)
\(loge\frac { N }{ { N }_{ 0 } } =-\lambda t\)
\(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
(b) (i) \(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept,if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process  
\(p\rightarrow n+{ e }^{ + }+\upsilon \)
(ii) Isobar

Give, at t = 0, number of nuclei of A = 4N₀ and number of nuclei of B = N₀.
Half-life of A, T_A = 1 min, half-life of B, T_B = 2 min
After time t, number of nuclei of A,
\({ n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/{ T }_{ A } }\)  
\(\Rightarrow \)  \({ n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/1 }\) 
After time t, number of nuclei of B,
 \({ n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/{ T }_{ A } } = { n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }\)   
Let the number of nuclei of A and B in given sample be after time t, then n_A = n_B
or \(4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/1 }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }\) or \({ \left( \frac { 1 }{ 2 } \right) }^{ t/2 }={ \left( \frac { 1 }{ 4 } \right) }={ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\) 
or \(\frac { t }{ 2 } =2\quad \Rightarrow \quad t=4min\) 
\(\therefore \ \ { n }_{ A }=4{ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 4/1 }=\frac { { N }_{ 0 } }{ 4 } \) 
and \({ n }_{ B }={ N }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 4/2 }=\frac { { N }_{ 0 } }{ 4 } \)  
\(\therefore \) Population of C in the sample is 
\(=\left( 4{ N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) +\left( \quad { N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) =\frac { { 9N }_{ 0 } }{ 2 } \)

\({ N }_{ 1 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 1 }t },\quad { N }_{ 2 }={ N }_{ 0 }{ e }^{ -{ \lambda }_{ 2 }t }\)
\(\\ \frac { { N }_{ 1 } }{ { N }_{ 2 } } ={ e }^{ \left( { \lambda }_{ 2 }-{ \lambda }_{ 1 } \right) t }\quad or\quad t=\frac { \log _{ e }{ \left( { N }_{ 1 }{ /N }_{ 2 } \right) } }{ { \lambda }_{ 2 }-{ \lambda }_{ 1 } } =\frac { \log _{ e }{ 140/1 } }{ \log _{ 2 }{ 2\left[ \frac { 1 }{ { T }_{ 2 } } -\frac { 1 }{ { T }_{ 1 } } \right] } } =\frac { \log _{ e }{ 140 } }{ \log _{ 10 }{ 2 } } \left( \frac { { T }_{ 2 }{ T }_{ 1 } }{ { T }_{ 1 }-{ T }_{ 2 } } \right) \)
On putting the values, we get \(t=6\times { 10 }^{ 9 }\ years\)