A) \[4\,\alpha ,\,\,1\beta \]
B) \[4\,\alpha ,\,\,2\beta \]
C) \[5\,\alpha ,\,\,1\beta \],
D) \[5\,\alpha ,\,\,2\beta \]
Correct Answer: A
Solution :
\[_{90}T{{h}^{228}}{{\xrightarrow{{}}}_{83}}B{{i}^{212}}+x_{2}^{4}He={{y}_{-1}}^{0}e\] Comparing mass numbers \[228=212+4x\] \[x=\frac{228-212}{4}=4\] Comparing the atomic number \[90=83+2x-y\] \[90=83+8-y\] \[y=91-90\] \[y=1\] Hence, number of a particles = 4 number of \[\beta \] particles = 1
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