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A mass spectrometer is a device which select particle of equal mass. An iron with electric charge $q > 0$ and mass $m$ starts at rest from a source $S$ and is accelerated through a potential difference $V$. It passes $\rho$ through a hole into a region of constant magnetic field $\vec B\,$ perpendicular to the plane of the paper as shown in the figure. The particle is deflected by the magnetic field and emerges through the bottom hole at a distance $d$ from the top hole. The mass of the particle is
$\frac{{qBd}}{{mV}}$
$\frac{{q{B^2}{d^2}}}{{4V}}$
$\frac{{q{B^2}{d^2}}}{{8V}}$
$\frac{{qBd}}{{2mV}}$
Solution
The speed of charged particle just before entering the magnetic field is $\mathrm{V}_{0}$.
$\mathrm{qV}=\frac{1}{2} \mathrm{mv}_{0}^{2}$
$v_{0}=\sqrt{\frac{2 q V}{m}}$
The radius of circular path in magnetic Field is
$r=\frac{m v_{0}}{q B}$
or $\frac{\mathrm{d}}{2}=\frac{\mathrm{mv}_{0}}{\mathrm{qB}}$
or $\frac{d^{2}}{4}=\frac{m^{2} v_{0}^{2}}{q^{2} B^{2}}=\frac{m^{2}}{q^{2} B^{2}} \frac{2 q V}{m}$
or $\frac{d^{2}}{4}=\frac{2 m V}{q B^{2}}$
$\therefore \quad \mathrm{m}=\frac{\mathrm{qB}^{2} \mathrm{d}^{2}}{8 \mathrm{V}}$
