Asked by Jane
The isotopic abundance of naturally occurring hydrogen is as follows:
1H 99.985 atom %
2H 0.015 atom%
When H2 gas is formed from naturally occurring hydrogen, what percentages of the molecules have molecular weights of approximately 2, 3 and 4?
1H 99.985 atom %
2H 0.015 atom%
When H2 gas is formed from naturally occurring hydrogen, what percentages of the molecules have molecular weights of approximately 2, 3 and 4?
Answers
Answered by
Graham
Compare to the tossing two coins.
P(Head ∩ Head) = P(Head) × P(Head) = 1/4
P(Head ∩ Tail) = 2 × P(Head) × P(Tail) = 1/2
P(Tail ∩ Tail) = P(Tail) × P(Tail) = 1/4
(Note that there are two ways to toss a head and a tail.)
The formation of hydrogen molecules from abundant hydrogen atoms is analogous, although there is a rather heavy bias against deuterium.
So:.....
P(<sup>1,1</sup>H<sub>2</sub>) = P(<sup>1</sup>H)×P(<sup>1</sup>H)
= 0.99985 × 99.985 %
=
P(<sup>1,2</sup>H<sub>2</sub>) = 2×P(<sup>1</sup>H)×P(<sup>2</sup>H)
= 2 × 0.99985 × 0.015 %
=
P(<sup>2,2</sup>H<sub>2</sub>) = P(<sup>2</sup>H)×P(<sup>2</sup>H)
= 0.00015 × 0.015 %
=
P(Head ∩ Head) = P(Head) × P(Head) = 1/4
P(Head ∩ Tail) = 2 × P(Head) × P(Tail) = 1/2
P(Tail ∩ Tail) = P(Tail) × P(Tail) = 1/4
(Note that there are two ways to toss a head and a tail.)
The formation of hydrogen molecules from abundant hydrogen atoms is analogous, although there is a rather heavy bias against deuterium.
So:.....
P(<sup>1,1</sup>H<sub>2</sub>) = P(<sup>1</sup>H)×P(<sup>1</sup>H)
= 0.99985 × 99.985 %
=
P(<sup>1,2</sup>H<sub>2</sub>) = 2×P(<sup>1</sup>H)×P(<sup>2</sup>H)
= 2 × 0.99985 × 0.015 %
=
P(<sup>2,2</sup>H<sub>2</sub>) = P(<sup>2</sup>H)×P(<sup>2</sup>H)
= 0.00015 × 0.015 %
=
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