Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hy

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Q. Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.

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Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Intuition First

Imagine you're holding a rope tied to a wall. If you flick your wrist once, a single pulse travels down the rope. If you flick faster — more frequently — each pulse carries more energy; the rope vibrates more violently. Light behaves the same way. A photon is the smallest possible "flick" of the electromagnetic field — a single, indivisible packet of light energy.

What determines how much energy that one photon carries? Two things: how fast the wave is oscillating (its frequency), and a universal constant that connects the wave world to the particle world.

The Precise Statement

The energy $E$ of a single photon is directly proportional to its frequency $f$ (or inversely proportional to its wavelength $\lambda$ ). The proportionality constant is Planck's constant $h$ , one of the most fundamental numbers in physics.

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (in joules, J)
$h$ = Planck's constant = $6.626 \times 10^{-34} \ \text{J·s}$
$f$ = frequency of the electromagnetic wave (in hertz, Hz)
$c$ = speed of light = $3.00 \times 10^{8} \ \text{m/s}$
$\lambda$ = wavelength of the light (in metres, m)

Why Two Forms?

The first form $E = hf$ is the most direct: higher frequency means higher energy. The second form $E = hc/\lambda$ is often more practical because wavelength is easier to measure than frequency. Since $c = f\lambda$ , you can always swap between them.

What This Tells You

Blue light (short wavelength, high frequency) has more energy per photon than red light (long wavelength, low frequency).
Gamma rays have enormous photon energies; radio waves have tiny photon energies.
The energy is quantised — you cannot have half a photon. Either the full energy $hf$ is absorbed/emitted, or none at all.

A Worked Example

Question: Calculate the energy of a single photon of violet light with wavelength $400 \ \text{nm}$ .

Step 1: Convert wavelength to metres.

400 \ \text{nm} = 400 \times 10^{-9} \ \text{m} = 4.00 \times 10^{-7} \ \text{m}

Step 2: Use $E = hc/\lambda$ .

E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^{8})}{4.00 \times 10^{-7}}

Step 3: Compute.

E = \frac{1.9878 \times 10^{-25}}{4.00 \times 10^{-7}} = 4.97 \times 10^{-19} \ \text{J}

Note

This is an incredibly tiny amount of energy — about $5 \times 10^{-19}$ joules. That's why we often use electronvolts (eV) for photon energies in atomic physics. $1 \ \text{eV} = 1.602 \times 10^{-19} \ \text{J}$ , so this photon has about $3.1 \ \text{eV}$ .

Common Mistake to Avoid …

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When an electron drops from $n=3$ to $n=2$ in hydrogen, it releases the energy difference between those levels as a photon. The energy is $1.89\,\text{eV}$ and the frequency is $4.57 \times 10^{14}\,\text{Hz}$ , corresponding to red light in the Balmer series.

Why this approach works

An electron in a hydrogen atom can only occupy certain discrete energy levels, each labeled by a principal quantum number $n$ . When the electron transitions from a higher level to a lower one, it must shed the exact energy difference, and it does so by emitting a photon. The photon's energy determines its frequency through Planck's relation $E = h\nu$ , and the energy levels themselves are given by the Rydberg formula for hydrogen.

The key insight: the photon carries away precisely $E_3 - E_2$ , the gap between the two states.

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

This formula gives the energy of an electron in the $n$ -th orbit of hydrogen, with the negative sign indicating that the electron is bound to the nucleus.

Step-by-step calculation

1. Find the energy of the initial state ( $n=3$ ).

E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51\,\text{eV}

2. Find the energy of the final state ( $n=2$ ).

E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.40\,\text{eV}

3. Calculate the energy released.

The energy of the emitted photon is the difference:

\Delta E = E_3 - E_2 = (-1.51) - (-3.40) = 1.89\,\text{eV}

The positive value confirms energy is released (emitted), not absorbed.

4. Convert the energy to joules (needed for frequency calculation).

\Delta E = 1.89 \times 1.6 \times 10^{-19}\,\text{J} = 3.024 \times 10^{-19}\,\text{J}

5. Calculate the frequency using Planck's relation.

$E = h\nu \quad \Rightarrow \quad \nu = \frac{E}{h}$ …

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Q.Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.

Why this approach works

Step-by-step calculation

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