Let the particle emitted in this can be represented as _zX^A.
Therefore, ₈₆R²²² \(\rightarrow \) ₈₄Po²¹⁸ + _zX^A.
Using the law of conservation of mass number and charge number, we get
222 = 218 + A and 86 = 84 + Z
\(\Rightarrow \)   A = 4 \(\Rightarrow \) Z = 2.
Now, A = 4 and Z = 2 correspond to \(\alpha \) - particle \(_{ 2 }^{ 4 }{ He }\) .
Therefore, emitted particle is \(\alpha \)-particle and the equation is ₈₆R²²² \(\rightarrow \) ₈₄Po²¹⁸+  \(_{ 2 }^{ 4 }{ He }\)

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 
\(=\left( 4{ N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) +\left( \quad { N }_{ 0 }-\frac { { N }_{ 0 } }{ 4 } \right) =\frac { { 9N }_{ 0 } }{ 2 } \)

Number of  -particles scattered at an angle of  is given by
\(N\alpha \frac { 1 }{ sin^{ 4 }\theta /2 } \Rightarrow \frac { { N }_{ 1 } }{ { N }_{ 2 } } =\left( \frac { \frac { \theta _{ 2 } }{ 2 } }{ sin\frac { { \theta }_{ 1 } }{ 2 } } \right) ^{ 4 }\)
\(\Rightarrow \ \frac { 100 }{ { N }_{ 2 } } =\left( \frac { sin\quad { 30 }^{ 0 } }{ sin\quad { 45 }^{ 0 } } \right) \Rightarrow \frac { 100 }{ { N }_{ 2 } } =\frac { 4 }{ 16 } \Rightarrow { N }_{ 2 }=400\)

Given, A = 238, R₀ = 1.2 x 10^-15 m
As, \(R={ R }_{ 0 }{ A }^{ { 1 }/{ 3 } }=1.2\times { 10 }^{ -15 }\left( { 238 }^{ { 1 }/{ 3 } } \right) \)
\(\therefore \)  \(R=7.437\times { 10 }^{ -15 }m\)

Number of protons, Z = 92
\(\therefore\) Number of neutrons, N = A-Z  = 238 - 92 = 146