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An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?

Respuesta :

hopemichael95

Answer:

3.9E-8

Explanation:

We know that

Mv²/r = Bqv

So

r= mv/Bq

But E is 1/2mv² which is 5.53eV

m²v² =2m x 5.53eV

mv = √( 2 x 9.1E-31)

So

r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19

= 3.9x10-8cm

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Explanation:

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