Question

The isotope chlorine-36 decays into argon-36, with a half-life of 3.01 × 105 years. Based on this information, fill in the blanks. A particular sample contains 3 × 1012 atoms of chlorine-36. After 9.03 × 105 years, there will × 10 atoms of chlorine-36 and × 10 atoms of argon-36.

Answer 1

A particular sample contains
`3* 10^(12)atoms`of chlorine-36. After
`9.03 * 10^5 years` , there will
`3.7533* 10^(11)` atoms of chlorine-36 and
`2.6246* 10^(12)`atoms of argon-36.

Step-by-step explanation:

Number of isotope chlorine-36 atoms =
`N_o=3* 10^(12) atoms`

Half-life of chlorine-36 atoms =
`t_{(1)/(2)}= 3.01* 10^5 years`

Number of chlorine-36 atom after time , t = N

t =
`9.03* 10^5 years`

`\lambda =(0.693)/(3.01* 10^5 years)`

`\log[N]=\log[N_o]-(\lambda * t)/(2.303)`

`\log[N]=\log[3* 10^(12) atoms]-(0.693* 9.03* 10^5 years)/(3.01* 10^5 years* 2.303)`

`N=3.7533* 10^(11) atoms`

Number of chlorine-36 atom after time =
`3.7533* 10^(11) atoms`

Number of argon-36 atom after time :

`=N_o-N=3* 10^(12) atoms=3.7533* 10^(11) atom`

`=2.6246* 10^(12) atoms`

Answer 2

After 9.03 × 105 years, there will 0.37 × 1012 atoms of chlorine-36 and 2.602 × 1012 atoms of argon-36.

Step-by-step explanation:

Half life an element is the time after which it decays to half of its initial concentration.

Half life of chlorine-36 is 3.01 * 105. And chlorine decays to Argon-36.

So 9.03 * 105 means, There are 3 half life of Chlorine.

• After 1st half life,

The amount of chlorine-36 = 1.5 * 1012

The amount of Argon-36 = 1.5 * 1012

• After 2nd half life:

The amount of chlorine-36 = 0.75 * 1012

The amount of Argon-36 = 2.25 * 1012

• After 3rd half life:

The amount of chlorine-36 = 0.372 * 1012

The amount of total Argon-36 = 2.602 * 1012