Given that
Potential of gun collector = 100 V
Magnetic fiels B = 2.83 * 10^{-4} T
Radius of the curved path r = 12.0 cm = 12 *10^2 m
Step 1: Determine the velocity of an ejected electron from gun due to voltage
Kinetic energy of electron K.E = \frac{1}{2} mv^2
v^2 = \frac{2 K.E}{m}
According to energy conversion K.E = energy of each electron = eV
v^2 = \frac{2 eV}{m} ....................(1)
Step 2: Velocity of electron in a spherical bulb due to magnetic field.
Electron moving in a circular path so it acquires centripetal force
F = \frac{mv^2}{r}
Force on the electron in a magnetic field
F =evB \because \theta = 90\degree
So, from the two equations we can write, evB = \frac{mv^2}{r}
eB = \frac{mv}{r}
v = \frac{eBr}{m}
v^2 = \frac{e^2B^2r^2}{m^2} ....................(2)
Step 3: Calculating the specific charge ratio
From equation (1) and (2)
\frac{e^2B^2r^2}{m} = \frac{2 eV}{m}
\frac{e}{m} = \frac{2V}{B^2 r^2}
Plug in the values in the above equation.
[math] \frac{e}{m} = \frac{2 * 100}{[2.83 * 10^{-4}]^2 (12*10^2)^2} [/math]
\frac{e}{m} = 1.734 * 10^{11} C/kg
Hence, the specific charge ratio \frac{e}{m} = 1.734 * 10^{11} C/kg