Answer
$1.8\times10^{11}C/kg$.
Work Step by Step
After the acceleration, the electron has a kinetic energy of 96 eV. Find the speed. Assume that the electron is nonrelativistic.
$$KE=\frac{1}{2}mv^2$$
$$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(96eV)(1.60\times10^{-19}J/eV)}{9.11\times10^{-31}kg}}$$
$$=5.807\times10^6 m/s$$
The electron is not relativistic.
In Example 20-6, the radius of the circular path is related to the mass, speed, charge, and the magnetic field strength. Solve for the charge-to-mass ratio.
$$r=\frac{mv}{qB}$$
$$\frac{q}{m}=\frac{v}{Br}$$
$$=\frac{5.807\times10^6 m/s }{(3.67\times10^{-4}T)(0.09m)}$$
$$= 1.8\times10^{11}C/kg $$