If 'f^{prime} denotes the ratio of the nu | vedclass

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Question

$N = N _{0} e ^{-\lambda t}$

$N _{ d }= N _{0}- N$

$N _{ d }= N _{0}\left(1- e ^{-\lambda t }\right)$

$\frac{ N _{ d }}{ N _{0}}= f =1- e ^{-

Vedclass · Accepted Answer

$\lambda e^{-\lambda t}$

Vedclass · Answer

$N = N _{0} e ^{-\lambda t}$

$N _{ d }= N _{0}- N$

$N _{ d }= N _{0}\left(1- e ^{-\lambda t }\right)$

$\frac{ N _{ d }}{ N _{0}}= f =1- e ^{-\lambda t }$

$\frac{ df }{ dt }=\lambda e ^{-\lambda t }$

If $'f^{\prime}$ denotes the ratio of the number of nuclei decayed $\left(N_{d}\right)$ to the number of nuclei at $t=0$ $\left({N}_{0}\right)$ then for a collection of radioactive nuclei, the rate of change of $'f'$ with respect to time is given as:

$[\lambda$ is the radioactive decay constant]

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If $'f^{\prime}$ denotes the ratio of the number of nuclei decayed $\left(N_{d}\right)$ to the number of nuclei at $t=0$ $\left({N}_{0}\right)$ then for a collection of radioactive nuclei, the rate of change of $'f'$ with respect to time is given as: $[\lambda$ is the radioactive decay constant]