What is the wavelength of a photon of energy $3\cdot3 \times 10^{-19}$ J ?

> [!TLDR] > The wavelength of a photon is found using the relation $E = \frac{hc}{\lambda}$. For $E = 3.3 \times 10^{-19}\ \text{J}$, the wavelength comes out to $6.0 \times 10^{-7}\ \text{m}$, which is 600 nm — in the visible (orange) region. The core idea here is the **photon energy–wavelength relation**. A photon carries energy in discrete packets, and that energy is inversely proportional to its wavelength. The constant of proportionality is the product of Planck’s constant $h$ and the speed of light $c$. This is one of the most fundamental equations in quantum physics — it connects the wave-like property (wavelength) with the particle-like property (energy). Whenever you see a problem asking for the wavelength of a photon given its energy, you reach straight for: > [!FORMULA] > $$E = \frac{hc}{\lambda}$$ where $h = 6.63 \times 10^{-34}\ \text{J·s}$ (Planck’s constant) $c = 3.0 \times 10^{8}\ \text{m/s}$ (speed of light in vacuum) $E$ is energy in joules, and $\lambda$ is wavelength in metres. We rearrange to solve for $\lambda$: $$\lambda = \frac{hc}{E}$$ Now let’s plug in the numbers carefully. 1. **Write down the given values** $E = 3.3 \times 10^{-19}\ \text{J}$ $h = 6.63 \times 10^{-34}\ \text{J·s}$ $c = 3.0 \times 10^{8}\ \text{m/s}$ 2. **Multiply $h$ and $c$** $$hc = (6.63 \times 10^{-34}) \times (3.0 \times 10^{8}) = 1.989 \times 10^{-25}\ \text{J·m}$$ > [!TIP] > The product $hc$ appears so often that it’s worth remembering: $hc \approx 1.99 \times 10^{-25}\ \text{J·m}$. In many problems, using $2.0 \times 10^{-25}$ is a quick approximation — but here we’ll keep the exact value for precision. 3. **Divide by the energy** $$\lambda = \frac{1.989 \times 10^{-25}}{3.3 \times 10^{-19}}$$ Handle the powers of ten first: $$\frac{10^{-25}}{10^{-19}} = 10^{-6}$$ So …

What is the wavelength of a photon of energy 3·3 × 10⁻¹⁹ J ?

Question

3\cdot3 \times 10^{-19}

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Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Core Idea

Light is not a smooth, continuous flow of energy — it comes in tiny indivisible packets called photons. Each photon carries a fixed amount of energy that depends only on the light's frequency (its colour), not on how bright the beam is. A brighter beam simply contains more photons; each individual photon still carries the same energy.

The Master Formula

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

where

$E$ = energy of one photon (joule, J),
$h = 6.63\times10^{-34}\ \text{J s}$ is Planck's constant,
$f$ = frequency of the light (hertz, Hz),
$c = 3\times10^8\ \text{m/s}$ is the speed of light, and
$\lambda$ = wavelength (metre, m).

The two forms are connected by the wave relation $c = f\lambda$ . Use $E = hf$ when you are given the frequency and $E = hc/\lambda$ when you are given the wavelength.

Note

Because $E = hc/\lambda$ , energy is inversely proportional to wavelength: short-wavelength radiation (X-rays, UV) has high-energy photons; long-wavelength radiation (radio, microwave) has low-energy photons.

Working in Electron-Volts

Photon energies are tiny in joules, so we often use the electron-volt:

1\ \text{eV} = 1.6\times10^{-19}\ \text{J}

A handy shortcut for visible/UV light expresses the energy directly from the wavelength in nanometres:

E(\text{eV}) \approx \frac{1240}{\lambda\ (\text{nm})}

(The number 1240 is just $hc$ expressed in eV·nm.)

Worked Example 1 — from frequency

Find the energy of a photon of frequency $f = 5.0\times10^{14}\ \text{Hz}$ (green light).

E = hf = (6.63\times10^{-34})(5.5\times10^{14}) = 3.6\times10^{-19}\ \text{J}

Converting to eV:

E = \frac{3.3\times10^{-19}}{1.6\times10^{-19}} \approx 2.1\ \text{eV}

Worked Example 2 — from wavelength

Find the energy of a photon of wavelength $\lambda = 620\ \text{nm}$ (red light).

E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^8)}{620\times10^{-9}} = 3.2\times10^{-19}\ \text{J} \approx 2.0\ \text{eV}

Or with the shortcut: $E \approx 1240/620 = 2.0\ \text{eV}$ — same answer, much faster.

Total Energy of a Beam

A single photon's energy is tiny, but a real beam contains enormous numbers of them. If a source emits $N$ photons per second (or a pulse contains $N$ photons), the total energy is simply

E_{\text{total}} = N \times hf

So the number of photons carrying a given power $P$ is …

Why this formula?

Photon Energy Calculation

Light of frequency $\nu$ (or wavelength $\lambda$ ) is carried in indivisible packets called photons. Calculating a photon's energy is one of the most common numerical tasks in modern physics, and it rests on a single relation.

A photon's energy depends only on its frequency (colour), not on how bright the beam is: $E = h\nu = \dfrac{hc}{\lambda}$ .

The Working Formula

E = h\nu = \frac{hc}{\lambda}

where $h = 6.63\times10^{-34}\ \text{J·s}$ (Planck's constant), $c = 3\times10^{8}\ \text{m/s}$ , $\nu$ is frequency (Hz) and $\lambda$ is wavelength (m). The two forms are linked by the wave relation $c = \nu\lambda$ , so $\nu = c/\lambda$ .

Two Handy Shortcuts

Product $hc$ : $hc = 6.63\times10^{-34}\times3\times10^{8} \approx 1.99\times10^{-25}\ \text{J·m}$ .
Energy in electron-volts (divide joules by $1.6\times10^{-19}$ ):

$E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}$ …

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The wavelength of a photon is found using the relation $E = \frac{hc}{\lambda}$ . For $E = 3.3 \times 10^{-19}\ \text{J}$ , the wavelength comes out to $6.0 \times 10^{-7}\ \text{m}$ , which is 600 nm — in the visible (orange) region.

The core idea here is the photon energy–wavelength relation. A photon carries energy in discrete packets, and that energy is inversely proportional to its wavelength. The constant of proportionality is the product of Planck’s constant $h$ and the speed of light $c$ .

This is one of the most fundamental equations in quantum physics — it connects the wave-like property (wavelength) with the particle-like property (energy). Whenever you see a problem asking for the wavelength of a photon given its energy, you reach straight for:

E = \frac{hc}{\lambda}

where

$h = 6.63 \times 10^{-34}\ \text{J·s}$ (Planck’s constant)

$c = 3.0 \times 10^{8}\ \text{m/s}$ (speed of light in vacuum)

$E$ is energy in joules, and $\lambda$ is wavelength in metres.

We rearrange to solve for $\lambda$ :

\lambda = \frac{hc}{E}

Now let’s plug in the numbers carefully.

Write down the given values

$E = 3.3 \times 10^{-19}\ \text{J}$

$h = 6.63 \times 10^{-34}\ \text{J·s}$

$c = 3.0 \times 10^{8}\ \text{m/s}$
Multiply $h$ and $c$

hc = (6.63 \times 10^{-34}) \times (3.0 \times 10^{8}) = 1.989 \times 10^{-25}\ \text{J·m}

Tip

The product $hc$ appears so often that it’s worth remembering: $hc \approx 1.99 \times 10^{-25}\ \text{J·m}$ . In many problems, using $2.0 \times 10^{-25}$ is a quick approximation — but here we’ll keep the exact value for precision.

Divide by the energy

\lambda = \frac{1.989 \times 10^{-25}}{3.3 \times 10^{-19}}

Handle the powers of ten first:

\frac{10^{-25}}{10^{-19}} = 10^{-6}

So …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set 55/2/11 mark
$3\cdot3 \times 10^{-19}$
›Reveal solutionSolution
The wavelength of a photon is found using the relation $E = \frac{hc}{\lambda}$ . For $E = 3.3 \times 10^{-19}\ \text{J}$ , the wavelength comes out to $6.0 \times 10^{-7}\ \text{m}$ , which is 600 nm — in the visible (orange) region.

The core idea here is the photon energy–wavelength relation. A photon carries energy in discrete packets, and that energy is inversely proportional to its wavelength. The constant of proportionality is the product of Planck’s constant $h$ and the speed of light $c$ .

This is one of the most fundamental equations in quantum physics — it connects the wave-like property (wavelength) with the particle-like property (energy). Whenever you see a problem asking for the wavelength of a photon given its energy, you reach straight for:
$E = \frac{hc}{\lambda}$
where

$h = 6.63 \times 10^{-34}\ \text{J·s}$ (Planck’s constant)

$c = 3.0 \times 10^{8}\ \text{m/s}$ (speed of light in vacuum)

$E$ is energy in joules, and $\lambda$ is wavelength in metres.

We rearrange to solve for $\lambda$ :
$\lambda = \frac{hc}{E}$
Now let’s plug in the numbers carefully.

Write down the given values

$E = 3.3 \times 10^{-19}\ \text{J}$

$h = 6.63 \times 10^{-34}\ \text{J·s}$

$c = 3.0 \times 10^{8}\ \text{m/s}$

Multiply $h$ and $c$

$hc = (6.63 \times 10^{-34}) \times (3.0 \times 10^{8}) = 1.989 \times 10^{-25}\ \text{J·m}$
Tip

The product $hc$ appears so often that it’s worth remembering: $hc \approx 1.99 \times 10^{-25}\ \text{J·m}$ . In many problems, using $2.0 \times 10^{-25}$ is a quick approximation — but here we’ll keep the exact value for precision.

Divide by the energy

$\lambda = \frac{1.989 \times 10^{-25}}{3.3 \times 10^{-19}}$
Handle the powers of ten first:
$\frac{10^{-25}}{10^{-19}} = 10^{-6}$
So …
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CBSE 2018Set ANNUAL1 markMCQ
$Q. The energy of an electromagnetic radiation is 13.2keV. This radiation is related to which region of spectrum?$
(a) Visible
(b) X-rays
(c) Ultra violet
(d) Infrared
›Reveal solutionSolution

keV-scale photon energies belong to X-rays.

Approximate photon energies by region: visible ≈ 1.6–3.3 eV, ultraviolet ≈ 3–100 eV, X-rays ≈ 100 eV to ~100 keV, gamma rays above that. Infrared is well below 1.6 eV.

An energy of 13.2 keV = 13,200 eV falls squarely in the X-ray band.

…

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Q.What is the wavelength of a photon of energy 3⋅3×10−193\cdot3 \times 10^{-19}3⋅3×10−19 J ?

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Q.What is the wavelength of a photon of energy $3\cdot3 \times 10^{-19}$ J ?

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