In nature, ratio of isotopes of Boron, 5 B 10 and 5 B 11, is (given that atomic weight of boron is 10.81 )

Question

In nature, ratio of isotopes of Boron, 5 B 10 and 5 B 11, is (given that atomic weight of boron is 10.81 ) (A) 81: 19 (B) 21: 44 (C) 19: 81 (D) 44: 21

Tardigrade Admin · Accepted Answer

Atomic weight =10.81 Weighted mean is hence =10.81 10.81=(10 × x+11 ×(100-x)/100) 1081=10 x+1100-11 x x=19 Hence, B10: B11=19: 81

Q. In nature, ratio of isotopes of Boron, $_{5} B^{10}$ and $_{5} B^{11}$ , is (given that atomic weight of boron is $10.81$ )

$81 : 19$

$21 : 44$

$19 : 81$

$44 : 21$

Atomic weight $= 10.81$
Weighted mean is hence $= 10.81$
$10.81 = \frac{10 \times x + 11 \times ( 100 - x )}{100}$
$1081 = 10 x + 1100 - 11 x$
$x = 19$
Hence, $B^{10} : B^{11} = 19 : 81$

Q. In nature, ratio of isotopes of Boron, 5​B10 and 5​B11, is (given that atomic weight of boron is 10.81 )

81:19

21:44

19:81

44:21