Question

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 this motion produces at the location of the proton.

Answer 1

Step-by-step explanation:

As the electron is revolving around the proton so, time taken in order to complete one revolution is the time period. Hence, formula for time period of the electron is as follows.

t =
`(2 \pi r)/(\\u)`

where, r = radius of the circle

`\\u` = speed of electron

Now, expression for current through the circular path is as follows.

I =
`(q)/(t)`

=
`(q\\u)/(2 \pi r)`

Now, magnetic field inside the circular path is as follows.

B =
`(\mu_(o)I)/(2r)`

or, =
`(\mu_(o))/(2r)((q\\u)/(2 \pi r))`

Putting the given values into the above formula as follows.

=
`(\mu_(o))/(2r)((q\\u)/(2 \pi r))`

=
`(4 \pi * 10^(-7) T A/m * 1.6 * 10^(-19) * 2.19 * 10^(6) m/s)/(4 \pi * 5.29 * 10^(-11))`

= 12.52 T

Thus, we can conclude that magnitude of the magnetic field is 12.52 T.