Answer:
a) the speed of the electron is 1.11 × 10⁷ m/s
b) the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m
Step-by-step explanation:
a) Let's assume that we have an electron accelerated using a potential difference of V = 350, which gives the ion a speed of v. In order to find this speed we set the potential energy of the electron equal to its kinetic energy. Thus,
eV = 1/2 m v²
where
- e is the charge of the electron
- m is the mass of the electron
- v is the speed of the electron
Thus,
v = √[2eV / m]
v = √[2(1.6 × 10⁻¹⁹ C)(350 V) / 9.11 × 10⁻³¹ kg]
v = 1.11 × 10⁷ m/s
Therefore, the speed of the electron is 1.11 × 10⁷ m/s
b) Then the electron enters a region of uniform magnetic field, it moves in a circular path with a radius of:
r = mv / eB
where
- m is the mass of the electron
- v is the speed of the electron
- e is the charge of the electron
- B is the magnetic field
Thus,
r = (9.11 × 10⁻³¹ kg)(1.11 × 10⁷ m/s) / (1.6 × 10⁻¹⁹ C)(200 × 10⁻³ T)
r = 3.16 × 10⁻⁴ m
Therefore, the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m