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In the Bohr model, the orbit n of a hydrogen-like ion has radius rn = 52.9 n²/Z pm, the angular…

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In the Bohr model, the orbit $n$ of a hydrogen-like ion has radius $r_n = 52.9\,\dfrac{n^2}{Z}\ \text{pm}$ , the angular momentum of the electron is $m_e v r = \dfrac{nh}{2\pi}$ , and its energy is $E_n = -2.18 \times 10^{-18}\,\dfrac{Z^2}{n^2}\ \text{J}$ . Consider the electron of a $\text{He}^{+}$ ion ( $Z = 2$ ) in the orbit $n = 3$ . Take $h = 6.626 \times 10^{-34}\ \text{J s}$ , $m_e = 9.109 \times 10^{-31}\ \text{kg}$ , $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$ and $c = 3.0 \times 10^{8}\ \text{m s}^{-1}$ .

What is the radius of this orbit?
How fast does the electron move in it?
What is its de Broglie wavelength?
What is its kinetic energy, in electronvolts?
The electron now falls straight to $n = 1$ . What is the wavelength of the light given out?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Radius from the formula, speed from the angular momentum, then the wavelength and the energy.

(a) Radius. $r_3 = 52.9 \times \dfrac{9}{2}\ \text{pm} = 238.05\ \text{pm}$ .

(b) Speed.

\displaystyle v = \frac{nh}{2\pi m_e r} = \frac{3 \times 6.626 \times 10^{-34}}{2\pi \times 9.109 \times 10^{-31} \times 2.3805 \times 10^{-10}}\ \text{m s}^{-1} = 1.459 \times 10^{6}\ \text{m s}^{-1}

(c) de Broglie wavelength.

\displaystyle \lambda = \frac{h}{m_e v} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1.459 \times 10^{6}}\ \text{m} = 498.6\ \text{pm}

The orbit is $2\pi r = 1495.7\ \text{pm}$ long, exactly three wavelengths. This is Bohr's rule seen as a standing wave: $n$ whole waves fit round orbit $n$ .

(d) Kinetic energy. $\tfrac12 m_e v^2 = 9.695 \times 10^{-19}\ \text{J} = 6.05\ \text{eV}$ . It equals the size of the orbit's energy, $2.18 \times 10^{-18} \times \tfrac49\ \text{J} = 9.689 \times 10^{-19}\ \text{J}$ , as the Bohr model says it must.

(e) The light. $\Delta E = 2.18 \times 10^{-18} \times 4 \times \left(1 - \dfrac19\right)\ \text{J} = 7.751 \times 10^{-18}\ \text{J}$ , so

\displaystyle \lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{7.751 \times 10^{-18}}\ \text{m} = 25.6\ \text{nm}

This is far in the ultraviolet.

Watch out. The de Broglie wavelength is not the wavelength of the light. One belongs to the moving electron, the other to the photon it gives out.

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