The half-life of a radioactive isotope is 4 | vedclass

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Question

(A) The number of half-lives $(n)$ is calculated as $n = \frac{	ext{Total time}}{	ext{Half-life}} = \frac{16 \ h}{4 \ h} = 4$.Remaining mass $(N_t)$

Vedclass · Accepted Answer

$187.5$

Vedclass · Answer

(A) The number of half-lives $(n)$ is calculated as $n = \frac{	ext{Total time}}{	ext{Half-life}} = \frac{16 \ h}{4 \ h} = 4$.Remaining mass $(N_t)$ is given by the formula $N_t = N_0 	imes (\frac{1}{2})^n$,where $N_0 = 200 \ g$.$N_t = 200 	imes (\frac{1}{2})^4 = 200 	imes \frac{1}{16} = 12.5 \ g$.The disintegrated amount is the initial mass minus the remaining mass: $	ext{Disintegrated amount} = N_0 - N_t = 200 \ g - 12.5 \ g = 187.5 \ g$.

The half-life of a radioactive isotope is $4 \ hours$. If the initial mass of the isotope is $200 \ g$,then what will be the disintegrated amount of it after $16 \ hours$?

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