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The natural boron is composed of two iso...

The natural boron is composed of two isotopes `._5beta^10` and `._5beta^11` having masses `10.003 u` and `11.009 u` resp. Find the relative abudance of each isotope in the natural boron if atomic mass of natural boron is 10.81u.

Text Solution

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Suppose natural boron contains x%
of `._5beta^10` and `(100-x)%` of `._5beta^11` isotope.
As atomic mass of natural boron =weigthed average of masses of two isotopes.
`:. 10.81 =(x xx 10.03+(100-x)11.009)/100`
`1081=-0.996x+1100.9`,
`x=19.98`.
Hence, relative abundance of `._5beta^10=19.98%`,
and relative abudance of `._5beta^11`
`100-19.98=80.02%`
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