Answer:
The speed of the electron is 6.79 x 10⁵ m/s
The radius of the circular path is 1.357 x 10⁻⁵ m
Explanation:
Given;
magnetic field, B = 0.285 T
energy of electron, E = 2.10 x 10⁻¹⁹ J
The kinetic energy of the electron is calculated as;
[tex]K.E = \frac{1}{2} m_eV^2[/tex]
Where;
[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg
V is the speed of the electron
[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]
The radius of the circular path is given by;
[tex]R = \frac{M_eV}{qB}[/tex]
where;
q is the charge of the electron = 1.6 x 10⁻¹⁹ C
[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]
