Question

In Niels Bohr's 1913 model of the hydrogen atom, an electron circles the proton at a...

In Niels Bohr's 1913 model of the hydrogen atom, an electron circles the proton at a distance of 5.29 ✕ 10-11 m with a speed of 2.19 ✕ 106 m/s. Compute the magnitude of the magnetic field that this motion produces at the location of the proton.

Homework Answers

Answer #1

Radius r = 5.29 x10 -11 m

Speed v = 2.19 x10 6 m/s

Orbital period of the electron T = 2r / v

T = 2(3.1415)(5.29 x10 -11 ) / (2.19 x10 6 )

     = 1.5177 x10 -16 s

Current setup by electron i = q/ T

i = (1.6 x10 -19 ) /(1.5177 x10 -16 )

    = 1.054 x10 -3 A

the magnitude of the magnetic field that this motion produces at the location of the proton i.e., magnetic field at the center of the loop B = i / 2r

                              = (4x10 -7 ) (1.054 x10 -3 ) /[2( 5.29 x10 -11)]

                              = 12.52 T

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