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x atoms of a radioactive element emit y beta particles per second . Then the decay constant of element is (in $\;s^{-1}$) .

$(a)\;xy\qquad(b)\;\large\frac{y}{x}\qquad(c)\;y^2 x\qquad(d)\;\large\frac{x}{y}$

1 Answer

Answer : (b) $\;\large\frac{y}{x}$
Explanation :
Activity of radioactive Substance $\;R=\lambda N$
$\lambda=\large\frac{R}{N}$
Here , R= y particles per second
and N=x
$\lambda=\large\frac{y}{x}$
answered Feb 28, 2014 by yamini.v
 

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