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On collision with neutron , .13Al^27 cha...

On collision with neutron , `._13Al^27` changes to radiosodium `._11Na^24` and emits a particle. `._11Na^24` , In its turn emits a particle and is transmitted to `._12Mg^24`. Write the two nuclear equations and identify the particles .

Text Solution

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let the first eqation of the reaction be
`._13Al^27 + ._0n^1 to ._11Na^24 + ._ZX^A`…(1)
and the second equation of the reaction be
`._11Na^24 to ._12Mg^24 + ._(Z_1)X^A_1` …(2)
Applying the laws of conservation of atomic number and mass number,
from equation (1), 27 + 1 = 24 + A , or , A=4
and 13+0 = 11 +Z or, Z=2
Also, from equation (2) , 24 = 24 + `A_1` or , `A_1`= 0
and `11=12 +Z_1` or , `Z_1=-1`
Hence, the particles are helium nucleus i.e., `alpha`-particle in equation (1) and `beta`-particle in equation (2).
The complete equations are `._13Al^27 +._0n^1 to ._11Na^24 + ._2He^4`
and `._11Na^24 to ._12Mg^24 + ._(-1)beta^0`
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