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

Explanation:

The given data is as follows.

  Half-life () for is year

or,             year

                = min

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

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

                           =

                           =

Since, it is given that mass of is 247.070 U. According to Avogadro's law, 247.070 g of contains atoms of .

Hence, number of atoms present in 0.00456 g of will be calculated as follows.

        0.00456 g of =

                                            = atoms of

It is known that expression for decay rate is

where,     = no. of atoms present in given amount of substance

Hence,       = 9.42 \times 10^{-10} min^{-1} \times 1.11 \times 10^{19}[/tex]

                                     =

As each decay is emitting only one alpha particle.

Therefore, we can conclude that number of alpha particles emitted per minute will be .

explanation:

i think option no d is the correct answer

a:

by the combination of an unlimited number of elements