A) \[6\alpha \,\text{and}\,\text{4}\beta \]
B) \[2\alpha \,\text{and}\,\text{4}\beta \]
C) \[6\alpha \,\text{and}\,2\beta \]
D) \[4\alpha \,\text{and}\,6\beta \]
Correct Answer: A
Solution :
\[{}_{90}T{{H}^{232}}\xrightarrow{{}}{}_{82}P{{b}^{208}}+x_{2}^{4}\alpha +{{y}_{1}}\beta \] |
Let the number of \[\alpha \]-particles be x. |
Let the number of \[\beta \]-particles be y. |
In radioactive disintegration atomic number as well as mass number must be equal to two sides of equation. |
\[\therefore \]\[208+4x+0y=232\] |
\[4x=24\] |
\[x=6\] |
and \[82+2x-y=90\] |
\[2x-y=8\Rightarrow y=4\] |
\[\therefore \]Number of \[\alpha \]-particles = 6 |
Number of \[\beta \]-particles = 4. |
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