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 \[\alpha \]particles = 4 number of \[\beta \] particles = 1
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