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13.Nuclei
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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
$81 : 19$
(AIPMT-1998)
Solution
(a) Let the percentage of ${B^{10}}$ atoms be $x,$ then Average atomic weight
$ = \frac{{10x + 11(100 – x)}}{{100}} = 10.81$
$ \Rightarrow x = 19\;\;\;\;$
$\therefore \frac{{{N_{{B^{10}}}}}}{{{N_{{B^{11}}}}}} = \frac{{19}}{{81}}$
Standard 12
Physics
