Question 11.22: 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 10-4 T 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 elm from the data.

 
Hint: \(evB=\frac{mv^2}{R}\)
Step 1: Find the velocity of the electron beam.

The energy of each electron is equal to its kinetic energy, i.e.,

12mv2=eVv2=2eVm                           

\( v=\sqrt{\frac{2eV}{m}}\)     .............. (1)
Step 2: Find the force acting on the electron beam and equate to centripetal force.
Force due to magnetic field = evB
Centripetal force = \(\frac{mv^2}{R}\)
\(evB=\frac{mv^2}{R}\\ \Rightarrow v=\frac{RBe}{m}........(2) \)
Step 3: Find the specific charge ratio.
From (1) & (2)
\(\begin{array}{l} R B\left[\frac{e}{m}\right]=\sqrt{2 V\left(\frac{e}{m}\right)} \\ R^{2} B^{2} \frac{e^{2}}{m^{2}}=2 v\left(\frac{e}{m}\right) \\ \left(\frac{e}{m}\right)=\frac{2 V}{R^{2} B^{2}} \end{array}\)
=2×100(2.83×104)2×(12×102)2=1.73×1011Ckg1

Therefore, the specific charge ratio (e/m) is 1.73 x 1011 Ckg-1