Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ion

Exercises · 2.10

kJ\ mol^{-1}

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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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The ionisation energy of sodium is found by converting the photon energy at the threshold wavelength (242 nm) from joules per atom to kilojoules per mole. The result is $494\ kJ\ mol^{-1}$ .

The key idea here is that "just sufficient to ionise" means the photon carries exactly enough energy to remove the most loosely bound electron from a sodium atom. That photon's energy is the ionisation energy per atom. We then scale it up to one mole using Avogadro's number.

Find the energy of a single photon. The energy of a photon is given by the Planck-Einstein relation:

E = \frac{hc}{\lambda}

where $h = 6.626 \times 10^{-34}\ J\ s$ (Planck's constant), $c = 3.0 \times 10^{8}\ m\ s^{-1}$ (speed of light), and $\lambda$ is the wavelength in metres.

Convert 242 nm to metres:

\lambda = 242\ nm = 242 \times 10^{-9}\ m = 2.42 \times 10^{-7}\ m

Plug in the numbers.

E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{2.42 \times 10^{-7}}

First, compute the numerator:

6.626 \times 10^{-34} \times 3.0 \times 10^{8} = 1.9878 \times 10^{-25}\ J\ m

Then divide by the wavelength:

E = \frac{1.9878 \times 10^{-25}}{2.42 \times 10^{-7}} = 8.214 \times 10^{-19}\ J

So each photon (and therefore each atom) requires $8.214 \times 10^{-19}\ J$ to ionise.

Tip

A useful shortcut: $hc = 1.9878 \times 10^{-25}\ J\ m$ is a constant you can memorise. For wavelength in nm, you can use $E = \frac{1240}{\lambda\ (nm)}\ eV$ if you want electronvolts, but here we need joules and then kilojoules per mole, so stick with SI units.

Convert from per atom to per mole. One mole contains $N_A = 6.022 \times 10^{23}$ atoms (Avogadro's number). The energy per mole is:

E_{mol} = (8.214 \times 10^{-19}\ J/atom) \times (6.022 \times 10^{23}\ atoms/mol)

Multiply: …

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Q.Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol−1kJ\ mol^{-1}kJ mol−1.

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Q.Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $kJ\ mol^{-1}$ .

Unlock everything free for 14 days