Question

Ordinary boron is a mixture of 10B5(mass no=10,atomic no=5)&11B5(mass no=11,atomic no=5)isotopes & has a composite atomic mass of 10.82u.What percentage of each isotope is present in ordinary boron?

Answer 1

`B_5^10: qquad 18"%"`

`B_5^11: qquad 82"%"`

Explanation:

`10 p + 11(1- p) = 10.82`

Solves as:

• `p = 9 /50 equiv 18"%"`

• `implies 1 - p = 41/50 equiv 82 "%"`