An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (–10^–2 mm of Hg). A magnetic field of 2.83 x 10^–4T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

Solution

Here,
Potential at collector, V = 100 V;
Magnetic field, B = 2.83 x 10^–4 T
Radius of the circular orbit, r = 12.0 cm = 12.0 x 10^–2 m
When electrons are accelerated through apotential of V volt, the gain in kinetic energy K.E. of the electron is given by,

$\frac{1}{2} {mv}^{2} = eV$

$\Rightarrow$    $v^{2} = \frac{2 eV}{m}$ ...(i)

Since the electron moves in circular orbit under magnetic field, therefore, force on the electron due to magnetic field provides the centripetal force to the electron.
$∴$ $evB = \frac{{mv}^{2}}{r}$

$\Rightarrow$    $eB = \frac{mv}{r}$

$\Rightarrow$    $v^{2} = \frac{e^{2} B^{2} r^{2}}{m^{2}}$ ....(ii)

From equations (i) and (ii), we get

$\frac{2 eV}{m} = \frac{e^{2} B^{2} r^{2}}{m^{2}}$

$\Rightarrow$    $\frac{e}{m} = \frac{2 V}{r^{2} B^{2}}$
$\frac{e}{m} = \frac{2 \times 100}{(12 \times 10^{- 2})^{2} \times (2.83 \times 10^{- 4})^{2}}$

$= 1.73 \times 10^{11} C {kg}^{- 1} .$

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