A)
B)
C)
D)
Correct Answer: D
Solution :
\[N={{N}_{0}}{{e}^{{{-}^{\lambda t}}}}\]and \[A={{A}_{0}}{{e}^{-\lambda t}}\]\[=\lambda {{N}_{0}}{{e}^{-\lambda t}}\] \ Ndecayed = N0 ? N = N0 ? N0e-ltÞ Ndecayed = \[{{N}_{0}}-\frac{A}{\lambda }\] This is equation of straight line with negative slope.
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