Nuclei Ques 6
6. Atomic weight of Boron is 10.81 and it has two isotopes $ _5 B^{10}$ and $ _5 B^{11}$. Then the ratio $ _5 B^{10}: _5 B^{11}$ in nature would be
[1998]
(a) $19: 81$
(b) $10: 11$
(c) $15: 16$
(d) $81: 19$
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Answer:
Correct Answer: 6.(a)
Solution:
- (a) Suppose that,
The number of ${ }^{10} B$ type atoms $=x$
and the number of ${ }^{11} B$ type atoms $=y$
Weight of ${ }^{10} B$ type atoms $=10 x$
Weight of ${ }^{11} B$ type atoms $=11 y$
Total number of atoms $=x+y$
$\therefore$ Atomic weight $=\frac{10 x+11 y}{x+y}=10.81$
$\Rightarrow 10 x+11 y=10.81 x+10.81 y$
$\Rightarrow 0.81 x=0.19 y \Rightarrow \frac{x}{y}=\frac{19}{81}$
If relative abundance of isotopes in an element has ratio $n_1: n_2$ whose atomic massers are $m_1$ and $m_2$ then atomic masses of elements is
$ M=\frac{n_1 m_1+n_2 m_2}{n_1+n_2} $
BETA
