Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and e

Exercises · 2.56

Q. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

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Concept understanding — Energy Level Quantization

Energy Level Quantization: From Intuition to Precision

Imagine you're climbing a smooth ramp. You can stop at any height — 1 metre, 1.5 metres, 2.1 metres — anywhere you like. That's how we intuitively think about energy in everyday life: continuous, like a slide.

Now imagine a staircase. You can stand on step 1, step 2, or step 3 — but you cannot stand halfway between step 2 and step 3. There's no such place. The steps are discrete, not continuous.

Energy level quantization is the idea that in the microscopic world of atoms and molecules, energy behaves like a staircase, not a ramp. Electrons in an atom cannot have just any energy — they can only occupy specific, allowed energy levels. Everything else is forbidden.

Why does this happen? The core intuition

In the classical world, an electron orbiting a nucleus would continuously radiate energy, spiral inward, and crash — atoms would be unstable. But atoms are stable. Nature solved this problem by imposing a rule: the electron's angular momentum (and therefore its energy) can only take certain discrete values.

Think of a guitar string. It can only vibrate at specific frequencies — its fundamental and harmonics. You can't pluck it to produce a frequency halfway between two harmonics. The string's vibration is quantized by its boundaries. Similarly, an electron bound to a nucleus is confined in space, and that confinement forces its energy to be quantized.

Note

Quantization is not a mysterious extra rule — it emerges naturally whenever a wave (like an electron's matter wave) is confined. Confinement creates standing waves, and standing waves only exist at specific frequencies.

The precise statement

For a bound system (like an electron in an atom), the total energy $E$ of the system can only take certain discrete values:

E = E_1, E_2, E_3, \dots

where each $E_n$ is a specific, fixed number. The integer $n$ (1, 2, 3, …) is called the principal quantum number. The lowest energy level ( $n=1$ ) is the ground state; higher levels ( $n>1$ ) are excited states.

For the hydrogen atom, the allowed energies are given by:

E_n = -\frac{13.6 \text{ eV}}{n^2}

So:

$n=1$ : $E_1 = -13.6$ eV (ground state)
$n=2$ : $E_2 = -3.4$ eV
$n=3$ : $E_3 = -1.51$ eV
and so on, approaching 0 eV as $n \to \infty$ (the ionization limit)

Important

The negative sign means the electron is bound to the nucleus. Zero energy corresponds to the electron being free (ionized). The more negative the energy, the more tightly bound the electron.

How do we know this is real?

The most direct evidence comes from atomic spectra. When an electron jumps from a higher energy level to a lower one, it emits a photon of light with energy exactly equal to the difference:

\Delta E = E_{\text{higher}} - E_{\text{lower}} = hf

where $h$ is Planck's constant and $f$ is the frequency of the emitted light. …

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The transition is from $n=5$ to $n=2$ in the hydrogen atom, giving a wavelength of 434.1 nm. This belongs to the Balmer series and lies in the visible region of the electromagnetic spectrum.

The key to solving this problem is recognising that the radii of Bohr orbits in hydrogen are quantised. Each orbit has a specific radius that depends only on the principal quantum number $n$ . Once we know the $n$ values for the initial and final orbits, we can use the Rydberg formula to find the wavelength of the emitted photon. The series and spectral region then follow directly from the $n$ values.

Let’s work through it step by step.

Recall the Bohr radius formula. For a hydrogen-like atom (here, hydrogen itself, $Z=1$ ), the radius of the $n$ -th orbit is given by:

r_n = n^2 a_0

where $a_0 = 52.9\ \text{pm}$ (the Bohr radius). This is a fundamental result: the radius scales as $n^2$ .

Watch out

A common mistake is to forget that the radius is proportional to $n^2$ , not $n$ . Doubling $n$ quadruples the radius.

Find the principal quantum number for the initial orbit. The initial radius is given as $1.3225\ \text{nm}$ . Convert to picometres for consistency:

1.3225\ \text{nm} = 1322.5\ \text{pm}

Using $r_n = n^2 a_0$ :

n_i^2 = \frac{r_i}{a_0} = \frac{1322.5\ \text{pm}}{52.9\ \text{pm}} = 25

So $n_i = \sqrt{25} = 5$ .

The electron starts in the $n=5$ orbit.

Find the principal quantum number for the final orbit. The final radius is $211.6\ \text{pm}$ . Again:

n_f^2 = \frac{211.6\ \text{pm}}{52.9\ \text{pm}} = 4

So $n_f = \sqrt{4} = 2$ .

The electron ends in the $n=2$ orbit.

Use the Rydberg formula to calculate the wavelength. For a hydrogen atom, the wavenumber $\tilde{\nu}$ (inverse wavelength) for a transition from $n_i$ to $n_f$ is:

\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

where $R_H = 1.097 \times 10^7\ \text{m}^{-1}$ (the Rydberg constant).

Here $n_i = 5$ , $n_f = 2$ :

\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{25} \right)

Compute the bracket: …

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Q.Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

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