Question

On average, an electron will exist in any given state in the hydrogen atom for about 10 − 8 s before jumping to a lower level. In the Bohr model of the hydrogen atom, estimate the number N of revolutions that an electron in the n = 3 energy level would make about the nucleus.

Answer 1

Answer:

θ = 4.1 rev

Explanation:

For this problem we can use the relationship of rotational kinematics

    w = θ / t

    θ = w t

Angular and linear variables are related.

     v = w r

     w = v/r

     θ = (v / r) t            (1)

We see that we must find the linear velocity and the radius of the orbit, let's use Bohr modeling

The radius of the orbit

          $r_{n}$ = a₀ n²

         a₀ = 0.0529 nm

For our case

n = 3

     r₃ = 0.0529 3²

     r₃ = 0.4761 nm

The energy of the atomic level is

    Eₙ = -13.606 / n²

n = 3

    E₃ = -13.606 / 3²

    E₃ = -1.512 eV

Let's reduce to July

   E₃ = -1.512 eV (1.6 10⁻¹⁹ J / 1 eV) = 2.4192 10⁻¹⁹ J

Let's use mechanical energy is

     E = K + U

     E = ½ m v² - k e² / r

     v² = (E + k e² /r)    2/m

     v² = (2.4192 10⁻¹⁹ + 8.99 10⁹ (1.6 10⁻¹⁹)² /0.4761 10-⁻⁹)     2/9.1 10⁻³¹

     v² = (2.4192 10⁻¹⁹ + 4.8339 10⁻¹⁹) 0.2198 10³¹

     v² = 1.5942 10¹²

     v = 1.23 10⁶ m / s

Let's replace in equation 1

     θ = v t / r

     θ = 1.23 10⁶  10⁻⁸ / 0.4761 10⁻⁹

     θ = 2.5835 10¹ rad

Let's reduce revolutions

    θ = 25,835 rad (1 rev / 2π rad)

    θ = 4.1 rev