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

63.750KeV

Step-by-step explanation:

We are given that

Initial velocity of second electron,
`u_2=0`

Radius,
`r_1=0`

`r_2=2.3 cm=(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* 10^(-31) kg`

Charge on an electron,
`q=-1.6* 10^(-19) C`

Velocity,
`v=(Bqr)/(m)`

Using the formula

Speed of electron,
`v_1=(Bqr_1)/(m)=(0.0370* 1.6* 10^(-19)* 0)/(9.1* 10^(-31))=0`

Speed of second electron,
`v_2=(0.0370* 1.6* 10^(-19)* 0.023)/(9.1* 10^(-31))`

`v_2=1.5* 10^8 m/s`

Kinetic energy of incident electron=
`(1)/(2)mv^2_1+(1)/(2)mv^2_2`

Kinetic energy of incident electron=
`0+(1)/(2)(9.1* 10^(-31))(1.5* 10^8)^2=1.02* 10^(-14) J`

Kinetic energy of incident electron=
`(1.02* 10^(-14))/(1.6* 10^(-19))=63750eV=(63750)/(1000)=63.750KeV`

1KeV=1000eV