Question

Measuring the approximate mass of an ion without using a standard can be accomplished via the following variant of the peak-matching technique described in Problem$20-14$. The peak-matching technique is used to alternately cause the $P^{+}$ion and the ${(P+1)}^{+}$ions to reach the detector. It is assumed that the difference in mass between $P^{+}$and ${(P+1)}^{+}$is due to a single ${}^{13}C$replacing a ${}^{12}C$atom.

(a) If the accelerating voltage for ${\left(P+1\right)}^{+}$is labeled $V_2$and that for $P^{+}$is $V_1$, derive a relationship that

relates the ratio $V_2/V_1$to the mass of $P^{+}$.

(b) If $V_2/V_1=0.987753$, calculate the mass of the $P^{+}$ion.

Short answer

The solutions for the following questions are :

(a) The relationship between the ratio $V_2/V_1$and the mass of $P^{+}$is $\frac{m\left(P^{+}\right)}{1.00335u+m\left(P^{+}\right)}=\frac{V_2}{V_1}$.

(b) Mass of the $P^{+}$ion is$80.92u$.

Step-by-step solution

Peak matching :

The rapid switching between the accelerating voltages when the other conditions are kept constant is called peak matching.

(a) The relationship between V2/V1 and mass P+ is obtained by :

The expression relating the ratio of accelerating voltages to corresponding $\frac{m}{z}$value is -

$\frac{{\left({\displaystyle \frac{m}{z}}\right)}_s}{{\left({\displaystyle \frac{m}{z}}\right)}_u}=\frac{V_u}{V_s}$

where,

Subscript $s$as standard value,

Subscript $u$as unknown value.

The mass difference expression between $P^{+}$and ${\left(P+1\right)}^{+}$is $m{(P+1)}^{+}-m\left(P^{+}\right)=m\left({}^{12}C\right)-m\left({}^{13}C\right)$

where mass of,

${\left(P+1\right)}^{+}$is $m{\left(P+1\right)}^{+}$,$P^{+}$is $m\left(P^{+}\right)$,

one ${}^{12}C$atom is $m\left({}^{12}C\right)$and

one ${}^{13}C$atom is $m\left({}^{13}C\right)$.

Substituting the values to find the relationship :

Substituting the values $13.00335u$for $m\left({}^{12}C\right)$and $12u$for $m\left({}^{13}C\right)$in the equation.

$$\begin{gathered} m{(P+1)}^{+}-m\left(P^{+}\right)=13.00335u-12u \\ m{(P+1)}^{+}-m\left(P^{+}\right)=1.00335u \\ m{(P+1)}^{+}=1.00335u+m\left(P^{+}\right) \end{gathered}$$

Substituting role="math" localid="1645636882277" $1.00335u+m\left(P^{+}\right)$for ${\left(\frac{m}{z}\right)}_u$,$V_2$for $V_u$,$V_1$for $V_s$and $m\left(P^{+}\right)$for ${\left(\frac{m}{z}\right)}_s$,

$\frac{m\left(P^{+}\right)}{1.00335u+m\left(P^{+}\right)}=\frac{V_2}{V_1}$

Hence, the equation shows the relationship.

(b) The mass of P+ ion can be calculated by : 

Substituting the given value of $\frac{V_2}{V_1}$as $0.987753$.

$\frac{m\left(P^{+}\right)}{1.00335u+m\left(P^{+}\right)}=0.987753$

$$\begin{gathered} m\left(P^{+}\right)=0.987753(1.00335u+m(P^{+}\left)\right) \\ m\left(P^{+}\right)=80.92u \end{gathered}$$

Hence, the mass is$80.92u$.