Respuesta :
Explanation:
The given data is as follows.
Half-life ([tex]t_{1/2}[/tex]) for [tex]^{247}Bk[/tex] is [tex]1.40 \times 10^{3}[/tex] year
or, [tex]1.40 \times 10^{3} \times 365 \times 24 \times 60 min[/tex] year
= [tex]7.36 \times 10^{8}[/tex] min
Relation between decay constant and half-life is as follows.
[tex]\lambda = \frac{0.693}{t_{1/2}}[/tex]
Therefore, putting the values into it we get the following.
[tex]\lambda = \frac{0.693}{t_{1/2}}[/tex]
= [tex]\frac{0.693}{7.36 \times 10^{8} min}[/tex]
= [tex]9.42 \times 10^{-10} min^{-1}[/tex]
Since, it is given that mass of [tex]^{247}Bk[/tex] is 247.070 U. According to Avogadro's law, 247.070 g of [tex]^{247}Bk[/tex] contains [tex]6.023 \times 10^{23}[/tex] atoms of [tex]^{247}Bk[/tex].
Hence, number of atoms present in 0.00456 g of [tex]^{247}Bk[/tex] will be calculated as follows.
0.00456 g of [tex]^{247}Bk[/tex] = [tex]\frac{6.023 \times 10^{23}}{247.070 g} \times 0.00456 g[/tex]
= [tex]1.11 \times 10^{19}[/tex] atoms of [tex]^{247}Bk[/tex]
It is known that expression for decay rate is [tex]\frac{-dN}{dt} = \lambda \times N_{o}[/tex]
where, [tex]N_{o}[/tex] = no. of atoms present in given amount of substance
Hence, [tex]\frac{-dN}{dt}[/tex] = 9.42 \times 10^{-10} min^{-1} \times 1.11 \times 10^{19}[/tex]
= [tex]1.045 \times 10^{10} min^{-1}[/tex]
As each decay is emitting only one alpha particle.
Therefore, we can conclude that number of alpha particles emitted per minute will be [tex]1.045 \times 10^{10} min^{-1}[/tex].
