Question

Atomic weight of boron is 10.81 and it has two isotopes $_{5}B^{10}\ and\ _{5}B^{11}$. Then ratio of $_{5}B^{10}\ :\ _{5}B^{11}$ in nature would be(a) 19 : 81(b) 10 : 11(c) 15 : 16(d) 81 : 19

Answer 1

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♦ Option [B] is Correct

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Answer 2

Answer:

a) 19:81

Explanation:

Atomic weight of boron = 10.81 (Given)

Isotope 1 = 5B10  (Given)

Isotope 2 = 5B11  (Given)

Thus, the required ratio -

Ratio = N1/N2

Average weight = N1W1+N2W2 / N1+N2

= 10.81 = 10N1 + 11N2/ N1+N2

= 10.81N1 = 10.81N2 = 10N1 + 11N2

= 0.81N1 = 0.19N2

= N1.N2 = 19/81

Thus, Then ratio of 5B10 and 5B11 in the nature will be 19:81