Is it possible to find the ratio of isotopes only given the mean mass number?

Question

Three isotopes of an element have mass numbers $(M)$, $(M+1)$ and $(M+2)$. If the mean mass number is $(M+0.5)$, then the ratio of the amounts of the three isotopes is?

Let amount of $(M)$, $(M+1)$, and $(M+2)$ be $x$, $y$ and $z$ respectively. $$\frac{(M)x + (M + 1)y + (M + 2)z}{x + y + z} = (M + 0.5)$$

After further simplification, I got $x-y=3z$.

I don't know how to proceed further. Can someone give me a hint, or reassure me that this question can't be solved with the given data?

The answer is given as $x:y:z=4:1:1$.

Answer 1

You have two equations and 3 unknowns, so you can't solve it with just that. Say a, b, c are the fractions (as a decimal) of each isotope...

$$ a(x) + b(x+1) + c(x+2) = (x+\frac{1}{2}) $$

$$a + b + c = 1 $$

The 4:1:1 solution works. Another that works is 3:0:1. Another is 7:4:1. There are infinitely many solutions.

Answer 2

As already stated, the system of equations is underdetermined. But we can get the range of possible solutions. Starting with \begin{equation} xM + y(M+1) + z(M+2) = M + \frac{1}{2} \end{equation} and using the normalization constraint \begin{equation} x + y + z = 1 \end{equation} we get \begin{equation} (x+y+z)M + y + 2z = M + \frac{1}{2} \\ y + 2z = \frac{1}{2} \end{equation} which can be rearranged to \begin{equation} z = \frac{1}{4} - \frac{y}{2} \\ y = \frac{1}{2} - 2z \end{equation} We can plug $z$ back into the normalization constraint and solve for $y$ \begin{equation} x + y + \frac{1}{4} - \frac{y}{2} = 1 \\ y = \frac{3}{2} - 2x \end{equation} which in turns allows us to express $z$ in terms of $x$ \begin{equation} z = \frac{1}{4} -\frac{1}{2}(\frac{3}{2} - 2x) \\ z = x - \frac{1}{2} \end{equation}

We further know that $x$, $y$ and $z$ are between 0 and 1. Therefore we can derive: \begin{equation} 0 \le y \le 1 \Rightarrow 0 \le \frac{3}{2} -2x \le 1 \\ \frac{3}{4} \ge x \ge \frac{1}{4} \end{equation} and similar from $z$ we get \begin{equation} \frac{1}{2} \le x \le \frac{3}{2} \end{equation} Combining all constraints we get all possible solutions for $x$: \begin{equation} \frac{1}{2}\le x \le \frac{3}{4} \end{equation}