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Question: The natural boron of atomic weight 10.81 is found to have tow isotopes \(10B\) and \(11B\) the ratio...

The natural boron of atomic weight 10.81 is found to have tow isotopes 10B10B and 11B11B the ratio of abundance of isotopes of natural boron should be

A

11 : 10

B

81 : 19

C

10 : 11

D

19 : 81

Answer

19 : 81

Explanation

Solution

: Let abundance of B10B^{10}be x%

∴\therefore Abundance of B11=(100−x)%B^{11} = (100 - x)\%

∴\therefore 10.81=(10×x)+11(100−x)10010.81 = \frac{(10 \times x) + 11(100 - x)}{100}

Or 1081=10x+1100−11x1081 = 10x + 1100 - 11xor x=19x = 19

∴\therefore Ratio of abundance

=19100−19=1981= \frac{19}{100 - 19} = \frac{19}{81}