Question

a sample of radioactive element contains 4×10^8 active nuclei . If the half life of elements ten days the no of decayed nuclei after 30 days is​

Answer 1

Given :

▪ No. of active nuclei = 4×10$^{10}$

▪ Half life of element = 10days

To Find :

▪ The number of decayed nuclei after 30 days.

SoluTion :

Number of half-lives

$\dashrightarrow\sf\:n=\dfrac{t}{T_{\frac{1}{2}}}=\dfrac{30\:days}{10\:days}\\ \\ \dashrightarrow\sf\:\red{n=3}$

[T_½ = half life period]

So, number of undecayed radioactive nuclei is given by

$\dashrightarrow\sf\:\dfrac{N}{N_o}=[\dfrac{1}{2}]^n$

• N = Final number
• No = Initial number

$\dashrightarrow\sf\:N=N_o[\dfrac{1}{2}]^n=4\times 10^{10}\times [\dfrac{1}{2}]^3\\ \\ \dashrightarrow\sf\:N=4\times 10^{10}\times \dfrac{1}{8}\\ \\ \dashrightarrow\sf\:\blue{N=0.5\times 10^{10}}$

Thus, number of nuclei decayed after 30 days is given by

$\dashrightarrow\sf\:N'=N_o-N\\ \\ \dashrightarrow\sf\:N'=(4\times 10^{10})-(0.5\times 10^{10})\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{N'=3.5\times 10^{10}}}}}\:\orange{\bigstar}$

Answer 2

$\large\underline\mathrm\red{Given:-}$

• $\large\mathrm{A \: sample \: of \: radioactive \: element \: contains \: 4×10^{8} \: active \: nuclei.}$

• $\large\mathrm{the \: half \: life \: of \: elements \: ten \: days \: the \: no \: of \: decayed \: nuclei \: after \: 30 \: days}$

$\large\underline\mathrm\red{Solution}$

$\large\mathrm{≫≫ N = no (1/2)^{t/T}}$

$\large\mathrm{≫≫ N = 4 × 10^{16} × (1/2)^{3}}$

$\large\mathrm{≫≫ N = 4 × 10^{16}}$

$\large\underline\mathrm\red{Now}$,

$\large\mathrm{≫≫ No \: of \: decayed \: atom = 4 × 10^{16} - 0.5 × 10^{16}</p><p>}$

$\large\mathrm{⟹ 3.5 × 10^{16}}$

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