Question

Assuming that each atom that decays emits one alpha particle, how many alpha particles are emitted per minute by a 0.00456-g sample of products? The half-life of 47Bk is 1.40x10°y and the mass of a 247Bk atom is 247,070 u. that is free from its decay, alpha particles min Submit Answer 5 question attempts remaining

Answer 1

Step-by-step explanation:

The given data is as follows.

Half-life (
`t_(1/2)`) for
`^(247)Bk` is
`1.40 * 10^(3)` year

or,
`1.40 * 10^(3) * 365 * 24 * 60 min` year

=
`7.36 * 10^(8)` min

Relation between decay constant and half-life is as follows.

`\lambda = (0.693)/(t_(1/2))`

Therefore, putting the values into it we get the following.

`\lambda = (0.693)/(t_(1/2))`

=
`(0.693)/(7.36 * 10^(8) min)`

=
`9.42 * 10^(-10) min^(-1)`

Since, it is given that mass of
`^(247)Bk` is 247.070 U. According to Avogadro's law, 247.070 g of
`^(247)Bk` contains
`6.023 * 10^(23)` atoms of
`^(247)Bk`.

Hence, number of atoms present in 0.00456 g of
`^(247)Bk` will be calculated as follows.

0.00456 g of
`^(247)Bk` =
`(6.023 * 10^(23))/(247.070 g) * 0.00456 g`

=
`1.11 * 10^(19)` atoms of
`^(247)Bk`

It is known that expression for decay rate is
`(-dN)/(dt) = \lambda * N_(o)`

where,
`N_(o)` = no. of atoms present in given amount of substance

Hence,
`(-dN)/(dt)` = 9.42 \times 10^{-10} min^{-1} \times 1.11 \times 10^{19}[/tex]

=
`1.045 * 10^(10) min^(-1)`

As each decay is emitting only one alpha particle.

Therefore, we can conclude that number of alpha particles emitted per minute will be
`1.045 * 10^(10) min^(-1)`.