An element containing only 2 isotopes , 50P and 53P (50 and 53 are the nucleon numbers) have been analysed in a mass spectrometer . If the relative atomic mass of P is 51.9 , what is the relative abundance of 50P ?
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If an element X has isotopes of $\displaystyle X_a \: , \: X_{b} \: , \: X_c \: ... \: X_n$ and relative abundance $\displaystyle n_a \: , \: n_{b} \: , \: n_c \: ... \: n_n$ respectively (note that the sum of relative abundances is always 1) then the relative atomic mass ($\displaystyle M_r$) is given by:
$\displaystyle M_r = {X_an_a + X_bn_b + X_cn_c + ... + X_nn_n}$
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As a rough check you can check the location of 50, 51.9 and 53 on a number line from 50 to 53. Note that 51.9 lies about 2/3 of the way along it thus it would follow that it will be roughly 1/3 of 50 and 2/3 of 53.
In your case
- $\displaystyle M_r = 51.9$
- $\displaystyle X_a = 50$
- $\displaystyle n_a = n_a$
- $\displaystyle X_b = 53$
- $\displaystyle n_b = (1-n_a)$
$\displaystyle 51.9 = 50n_a + 53 - 53n_a$
Solve for $\displaystyle n_a$. Express in whichever method your teacher wants.
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