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In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius $R$ using a magnetic field $B$. If $V$ and $B$ are kept constant, the ratio $\left(\frac{\text { charge on the ion }}{\text { mass of the ion }}\right)$ will be proportional to
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The correct answer is:
$\frac{1}{R^2}$
The radius of the orbit in which ions moving is determined by the relation as given below
$\frac{m v^2}{R}=q v B$
where $m$ is the mass, $v$ is velocity, $q$ is charge of ion and $B$ is the flux density of the magnetic field, so that $q v B$ is the magnetic force acting on the ion, and $\frac{m v^2}{R}$ is the centripetal force on the ion moving in a curved path of radius $R$.
The angular frequency of rotation of the ions about the vertical field $B$ is given by
$\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v$
where $v$ is frequency.
Energy of ion is given by
or
$\begin{aligned}
E & =\frac{1}{2} m v^2 \\
& =\frac{1}{2} m(R \omega)^2 \\
& =\frac{1}{2} m R^2 B^2 \frac{q^2}{m^2} \\
E & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \ldots(i)
\end{aligned}$
If ions are accelerated by electric potential $V$, then energy attained by ions
$E=q V\ldots(ii)$
From Eqs. (i) and (ii), we get
$\begin{aligned}
q V & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \\
or\frac{q}{m} & =\frac{2 V}{R^2 B^2}
\end{aligned}$
If $V$ and $B$ are kept constant, then
$\frac{q}{m} \propto \frac{1}{R^2}$
$\frac{m v^2}{R}=q v B$
where $m$ is the mass, $v$ is velocity, $q$ is charge of ion and $B$ is the flux density of the magnetic field, so that $q v B$ is the magnetic force acting on the ion, and $\frac{m v^2}{R}$ is the centripetal force on the ion moving in a curved path of radius $R$.
The angular frequency of rotation of the ions about the vertical field $B$ is given by
$\omega=\frac{v}{R}=\frac{q B}{m}=2 \pi v$
where $v$ is frequency.
Energy of ion is given by
or
$\begin{aligned}
E & =\frac{1}{2} m v^2 \\
& =\frac{1}{2} m(R \omega)^2 \\
& =\frac{1}{2} m R^2 B^2 \frac{q^2}{m^2} \\
E & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \ldots(i)
\end{aligned}$
If ions are accelerated by electric potential $V$, then energy attained by ions
$E=q V\ldots(ii)$
From Eqs. (i) and (ii), we get
$\begin{aligned}
q V & =\frac{1}{2} \frac{R^2 B^2 q^2}{m} \\
or\frac{q}{m} & =\frac{2 V}{R^2 B^2}
\end{aligned}$
If $V$ and $B$ are kept constant, then
$\frac{q}{m} \propto \frac{1}{R^2}$
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