Question

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories are 0.00 cm and 2.30 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0370 T. Determine the energy (in keV) of the incident electron.

Answer 1

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron,$u_2=0$

Radius,$r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron,$m=9.1\times 10^{-31} kg$

Charge on an electron,$q=-1.6\times 10^{-19} C$

Velocity,$v=\frac{Bqr}{m}$

Using the formula

Speed of electron,$v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron,$v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron=$\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron=$0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron=$\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV