Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as $\nu = 3.29 \times 10^{15}\ (Hz)\ \left[\dfrac{1}{3^2} - \dfrac{1}{n^2}\right]$. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as

Exercises · 2.55

\nu = 3.29 \times 10^{15}\ (Hz)\ \left[\dfrac{1}{3^2} - \dfrac{1}{n^2}\right]

Meghalaya MboseTextbookSubjective· 3mImportance★★★★★est

52% · 73/140 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

The Paschen series transition at 1285 nm corresponds to a jump from $n=5$ to $n=3$ , placing it in the infrared region of the electromagnetic spectrum.

The key to this problem is understanding that the Paschen series is defined by transitions that end at the third energy level ( $n=3$ ). The formula given is the Rydberg formula for hydrogen, where the constant $3.29 \times 10^{15}$ Hz is the Rydberg frequency ( $R_H c$ ). When a problem gives you a wavelength, your first instinct should be to convert it to frequency, because the formula is written in terms of frequency ( $\nu$ ).

Let's walk through this step by step.

Convert wavelength to frequency. The relationship between frequency ( $\nu$ ), wavelength ( $\lambda$ ), and the speed of light ( $c$ ) is:

\nu = \frac{c}{\lambda}

Here, $\lambda = 1285\ \text{nm} = 1285 \times 10^{-9}\ \text{m}$ . Using $c = 3.00 \times 10^8\ \text{m/s}$ :

\nu = \frac{3.00 \times 10^8}{1285 \times 10^{-9}} = \frac{3.00 \times 10^8}{1.285 \times 10^{-6}}

\nu = 2.334 \times 10^{14}\ \text{Hz}

Set up the Rydberg equation for the Paschen series. The given formula is:

\nu = 3.29 \times 10^{15} \left[ \frac{1}{3^2} - \frac{1}{n^2} \right]

Substitute the frequency we found:

2.334 \times 10^{14} = 3.29 \times 10^{15} \left[ \frac{1}{9} - \frac{1}{n^2} \right]

Isolate the bracket term. Divide both sides by $3.29 \times 10^{15}$ :

\frac{2.334 \times 10^{14}}{3.29 \times 10^{15}} = \frac{1}{9} - \frac{1}{n^2}

0.07094 \approx \frac{1}{9} - \frac{1}{n^2}

Solve for $1/n^2$ . Since $\frac{1}{9} \approx 0.1111$ , we have:

0.07094 = 0.1111 - \frac{1}{n^2}

\frac{1}{n^2} = 0.1111 - 0.07094 = 0.04016

Find $n$ . Take the reciprocal:

n^2 = \frac{1}{0.04016} \approx 24.90

n \approx \sqrt{24.90} \approx 4.99

Since $n$ must be an integer (energy levels are discrete), $n = 5$ . …

Start your 14-day free trial to unlock the full solution →

←Previous question · exercises2.54 — If the photon of the wavelength 150 pm strikes an atom and one of its …→Next question · exercises2.56 — Calculate the wavelength for the emission transition if it starts from…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days