Calculate the mass of 1 curie of R_aB (Pb²) from its half-life of 26.8 minutes.

Solution

Given,
Half life of RaB = 26.8 min

Activity of $R_{a} B = \frac{dN}{dt} = 1 curie = 3.7 \times 10^{10} disintegrations / s$

If, $λ$ is the disintegration constant of $R_{a} B, we have$
$λ = \frac{0.693}{T}$

$= \frac{0.693}{26.8 \times 60} s^{- 1}$

Let N be the number of atoms of $R_{a} B$ having an activity of 1 curie. Then ,

we have    $|\frac{dN}{dt}| = λN$

$\Rightarrow$     $N = \frac{|\frac{dN}{dt}|}{λ}$

$∴$    $N = \frac{3.7 \times 10^{10} \times 26.8 \times 60}{0.693}$

Further we know that,

1g of atom = 6.02 $\times$ 10²³ atoms.

Mass of 1 curie of atom = 214g = $\frac{214}{6.02 \times 10^{23}}$

Therefore, the mass of $R_{a} B$ having an activity of 1 curie is,

$= \frac{214 \times 3.7 \times 10^{10} \times 26.8 \times 60}{6.02 \times 10^{23} \times 0.693} = 30.52 \times 10^{- 9} g .$

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Calculate the mass of 1 curie of R_aB (Pb²) from its half-life of 26.8 minutes.

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Wave Optics

Calculate the mass of 1 curie of RaB (Pb2) from its half-life of 26.8 minutes.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Calculate the mass of 1 curie of R_aB (Pb²) from its half-life of 26.8 minutes.