Let Nβ be the number of β particles emitted by 1 gram of Na24 radioactive nuclei (half life = 15 hrs) in 7.5 hours, Nβ is close to (Avogadro number = 6.023 × 1023 / g mole) :

Question

Let Nβ be the number of β particles emitted by 1 gram of Na24 radioactive nuclei (half life = 15 hrs) in 7.5 hours, Nβ is close to (Avogadro number = 6.023 × 1023 / g mole) : (A) 6.2 × 1021 (B) 7.5 × 1021 (C) 1.25 × 1022 (D) 1.75 × 1022

Tardigrade Admin · Accepted Answer

N=N0 e-λ t=No((1/2))x Where x= Number of half lives Here x=(7.5/15)=(1/2) h, N= (No/√2) ⇒ N=(6.023 × 1023/24 √2)=1.75 × 1022

Q. Let $N_{β}$ be the number of $β$ particles emitted by $1 g r am$ of $N a^{24}$ radioactive nuclei (half life = $15 h rs$ ) in $7.5$ hours, $N_{β}$ is close to (Avogadro number = $6.023 \times 1 0^{23}$ / g mole) :

$6.2 \times 1 0^{21}$

$7.5 \times 1 0^{21}$

$1.25 \times 1 0^{22}$

$1.75 \times 1 0^{22}$

$N = N_{0} e^{- λ t} = N_{o} (\frac{1}{2})^{x}$
Where $x =$ Number of half lives
Here $x = \frac{7.5}{15} = \frac{1}{2} h, N = \frac{N _{o}}{2}$
$\Rightarrow N = \frac{6.023 \times 1 0 ^{23}}{24 2} = 1.75 \times 1 0^{22}$

Q. Let Nβ​ be the number of β particles emitted by 1gram of Na24 radioactive nuclei (half life = 15hrs) in 7.5 hours, Nβ​ is close to (Avogadro number = 6.023×1023 / g mole) :

6.2×1021

7.5×1021

1.25×1022

1.75×1022