In the Bohr model of the hydrogen atom, an electron in the 2nd excited state moves at a speed of 5.48 105 m/s in a circular path of radius 2.12 10-10 m. What is the effective current associated with this orbiting electron

Respuesta :

Answer:

The current is [tex]I = 6.59*10^{-5} \ A[/tex]

Explanation:

From the question we are told that

  The speed is  [tex]v = 5.48*10^{5} \ m/s[/tex]

   The radius is [tex]r = 2.12 *10^{-10} \ m[/tex]

Generally the period of the orbiting electron is mathematically represented as

        [tex]T = \frac{2 \pi * r }{v}[/tex]    

=>    [tex]T = \frac{2 * 3.142 * 2.12 *10^{-10} }{ 5.48 *10^{5}}[/tex]    

=>    [tex]T = 2.43 *10^{-15} \ s[/tex]

Gnerally the effective current associated with this orbiting electron

     [tex]I = \frac{e}{T}[/tex]

Here e is the charge on an electron with value  [tex]e = 1.60 *10^{-19} \ C[/tex]

So

      [tex]I = \frac{1.60 *10^{-19}}{2.43 *10^{-15}}[/tex]

=> [tex]I = 6.59*10^{-5} \ A[/tex]

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