question_answer

                Atomic weight of boron is 10.81 and it has two isotopes \[_{5}^{10}B\] and \[_{5}^{11}B.\] Then, the ratio of atoms of \[_{5}^{10}B\] and \[_{5}^{11}B\] in nature, would be:

A)                                                                                                                                            19 : 81                 Let \[{{n}_{1}}\] and \[{{n}_{2}}\] be the number of atoms in \[_{5}^{10}B\] and \[_{5}^{11}B\] isotopes.                 Atomic weight                 \[=\frac{{{n}_{1}}\times (At.\,wt.\,of\,_{5}^{10}B)+{{n}_{2}}\times (At.wt.\,of\,_{5}^{11}B)}{{{n}_{1}}+{{n}_{2}}}\] \[or10.81=\frac{{{n}_{1}}\times 10+{{n}_{2}}\times 11}{{{n}_{1}}+{{n}_{2}}}\] \[or10.81\,{{n}_{1}}+10.81\,{{n}_{2}}=10\,{{n}_{1}}+11\,{{n}_{2}}\] \[or0.81\,\,{{n}_{1}}=0.19\,{{n}_{2}}\] \[or\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{0.19}{0.81}=\frac{19}{81}\]

B)                 10 : 11  

C)                 15 : 16 

D)                 81 : 19