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Charra [1.4K]
3 years ago
9

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.
Physics
1 answer:
morpeh [17]3 years ago
3 0

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

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