Exercise 26.23 Part A In a microwave oven, electrons describe circular motion in
ID: 1787799 • Letter: E
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Exercise 26.23 Part A In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called a magnetron; as you'll learn in Chapter 29 the electrons' motion results in the production of micowaves. The electrons circle at a frequency of 2.35 GHz. The magnetron can accommodate electron orbits with maximum diameter 2.92 mm What's the magnetic field strength? mT Submit My Answers Give Up Part B What's the electrons' energy in eV Submit My Answers Give Up de Feedback ContinueExplanation / Answer
Let the mass of the electron be m = 9.109 × 10-31 kg
Let the charge of the electron be q = 1.6 × 10-19 C
Let the magnetic field be B
Let the frequency of the electron be f = 2.35 GHz = 2.35 × 109 Hz
a) The relation between frequency and magnetic field is
f = B q / ( 2 m )
B = ( 2 m f ) / q
= ( 2 × 3.14 × 9.109 × 10-31 × 2.35 × 109 ) / ( 1.6 × 10-19 )
= ( 134.43 × 10-3 ) /1.6
= 84.019 × 10-3 T
So magnetic field is B = 84.019 × 10-3 T = 84.019 mT
B) To find the energy of the electron , we need to first find its velocity.
Velocity of the electron is given by V = q B r / m
Where r is radius of the circular path traced by the electron.
Diameter of the circular path traced by the electron D = 2.92 mm
So radius is r =. D / 2 = 2.92 / 2 = 1.46 mm = 1.46 × 10-3 m
Now velocity V = q B r / m
= ( 1.6 × 10-19 × 84.019 × 10-3 × 1.46 × 10-3 ) / ( 9.109 × 10-31 )
= 21.54 × 106 m/s
Now kinetic energy of the electron is E = 1/2 m V2
= 1/2 × ( 9.109 × 10-31 × 21.54 × 106 × 21.54 × 106 )
= 2113.15 × 10-19 J
Now let us calculate the energy in electron volts
1 eV = 1.6 × 10-19 J . So 1 joule = 1 / ( 1.6 × 10-19 ) eV
Then the energy E = ( 2113.15 × 10-19 ) / ( 1.6 × 10-19 )
= 1320.72 eV
Thus the energy of the electron is E = 1320.72 eV
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