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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.

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Answer:

θ = 4.1 rev

Step-by-step 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

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