Starting with a sample of pure {}^{66}Cu | vedclass

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Question

$\frac{7}{8}$ days of Cu decays.

$	herefore $ Cu undecayed, $\mathrm{N}=1-\frac{7}{8}=\frac{1}{8}=\left(\frac{1}{2}ight)^{3}$

$	herefore $ N

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$\frac{7}{8}$ days of Cu decays.

$	herefore $ Cu undecayed, $\mathrm{N}=1-\frac{7}{8}=\frac{1}{8}=\left(\frac{1}{2}ight)^{3}$

$	herefore $ No. of half lifes $=3$

$\mathrm{n}=\frac{\mathrm{t}}{\mathrm{T}}$ or $3=\frac{15}{\mathrm{T}}$

$\Rightarrow$ half life period, $\mathrm{T}=\frac{15}{3}=5$ $minutes$

Starting with a sample of pure ${}^{66}Cu,\frac{7}{8}$ of it decays into $Zn$ in $15\, minutes$. The corresponding half life is..........$minutes$

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