13.10 The half-life of <math><mmultiscripts><mrow><mrow><mi mathvariant="bold-italic">S</mi><mi mathvariant="bold-italic">r</mi></mrow></mrow><mprescripts></mprescripts><mrow><mrow><mn>38</mn></mrow></mrow><mrow><mrow><mn>90</mn></mrow></mrow></mmultiscripts></math> is 28 years. What is the disintegration rate of 15 mg of this isotope?

13.10 Half life of Sr3890 , T1/2 = 28 years = 28 ×365×24×60×60 secs = 0.883 ×109 sMass of the isotope, m = 15 mg = 15 ×10-3 gms90 g of Sr3890 contains 6.023 ×1023 atomsNo. of atoms in 15 mg of Sr3890 contains = 6.023×102390× 15 ×10-3 = 1.0038 ×1020Rate of disintegration dNdt = ?N , where ? = 0.693T1/2 = 0.6930.883×109 /s = 7.848 ×10-10 s-1dNdt = 7.848 ×10-10× 1.0038 ×1020 = 7.878 ×1010 atoms / second.

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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

13.10 The half-life of ${}_{38}^{90}{S r}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

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Payal Gupta | Contributor-Level 10

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13.10 Half life of ${}_{38}^{90}{S r}$ , $T_{1 / 2}$ = 28 years = 28 $* 365 * 24 * 60 * 60$ secs = 0.883 $* 10^{9}$ s

Mass of the isotope, m = 15 mg = 15 $* 10^{- 3}$ gms

90 g of ${}_{38}^{90}{S r}$ contains 6.023 $* 10^{23}$ atoms

No. of atoms in 15 mg of ${}_{38}^{90}{S r}$ contains = $\frac{6.023 * 10^{23}}{90} *$ 15 $* 10^{- 3}$ = 1.0038 $* 10^{20}$

Rate of disintegration $\frac{d N}{d t}$ = $? N$ , where $?$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{0.883 * 10^{9}}$ /s = 7.848 $* 10^{- 10}$ $s^{- 1}$

$\frac{d N}{d t}$ = 7.848 $* 10^{- 10} *$ 1.0038 $* 10^{20}$ = 7.878 $* 10^{10}$ atoms / second.

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