One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimu

NCERT Exemplar · Q1

Q. One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

(a) visible region.

(b) infrared region.

(d) microwave region.

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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

N = \frac{P}{hf} = \frac{P\lambda}{hc}\quad \text{(photons per second)}.

Watch out

Keep units consistent: put $\lambda$ in metres and $f$ in hertz before substituting, unless you are deliberately using the $1240/\lambda(\text{nm})$ eV shortcut.

Summary

One photon's energy: $E = hf = hc/\lambda$ .
Higher frequency (shorter wavelength) $\Rightarrow$ more energetic photon.
Convert to eV using $1\ \text{eV}=1.6\times10^{-19}\ \text{J}$ , or use $E(\text{eV})=1240/\lambda(\text{nm})$ .
A beam of power $P$ delivers $N = P/hf$ photons per second.

Mastering this calculation is the key to problems on the photoelectric effect, spectra, radiation energy and the whole quantum picture of light.

Calculating photon energy using E = hf = hc/λ is a foundational numerical skill from the NCERT Class 12 Physics chapter on dual nature of radiation and matter, tested heavily in CBSE boards, JEE Main and NEET. Searches for "photon energy formula numericals class 12 physics important questions" will find this eV-conversion and wavelength-based approach matches the NCERT-prescribed method.

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})}

This is the fastest route in exams when the wavelength is given in nanometres.

Worked Idea

Find the energy of a photon of green light, $\lambda = 500\ \text{nm} = 500\times10^{-9}\ \text{m}$ .

E = \frac{hc}{\lambda} = \frac{1.99\times10^{-25}}{500\times10^{-9}} = 3.98\times10^{-19}\ \text{J}

In electron-volts, $E = \dfrac{1240}{500} \approx 2.48\ \text{eV}$ .

Tip

Halving the wavelength doubles the photon energy; increasing the intensity only sends more photons, each still of energy $h\nu$ .

The energy required to dissociate CO is 11 eV. Using $E = h\nu$ , the minimum frequency is $\nu = E/h$ . Converting 11 eV to joules and dividing by Planck’s constant gives $\nu \approx 2.66 \times 10^{15} \, \text{Hz}$ , which lies in the ultraviolet region of the electromagnetic spectrum.

The core idea here is photon energy calculation. When electromagnetic radiation interacts with a molecule, each photon carries a discrete amount of energy given by $E = h\nu$ , where $h$ is Planck’s constant and $\nu$ is the frequency. For dissociation to occur, a single photon must supply at least the bond energy — in this case, 11 eV. If the photon’s energy is less, the molecule won’t break apart, no matter how many photons you throw at it. So the minimum frequency corresponds exactly to the photon energy equalling the dissociation energy.

The trick is to work in consistent units. The dissociation energy is given in electronvolts (eV), a convenient unit for atomic-scale energies, but Planck’s constant is usually given in joule-seconds. So we need to convert.

Convert the energy from eV to joules. One electronvolt is $1.602 \times 10^{-19}$ J. Therefore:

E = 11 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 1.7622 \times 10^{-18} \, \text{J}.

Apply the photon energy relation. The minimum frequency $\nu_{\text{min}}$ satisfies $E = h \nu_{\text{min}}$ , so:

\nu_{\text{min}} = \frac{E}{h}.

Planck’s constant $h = 6.626 \times 10^{-34} \, \text{J·s}$ . Substituting:

\nu_{\text{min}} = \frac{1.7622 \times 10^{-18}}{6.626 \times 10^{-34}} \approx 2.66 \times 10^{15} \, \text{Hz}.

Identify the spectral region.

The electromagnetic spectrum is divided roughly as:
- Radio: $< 10^9$ Hz
- Microwave: $10^9$ – $10^{12}$ Hz
- Infrared: $10^{12}$ – $4 \times 10^{14}$ Hz
- Visible: $4 \times 10^{14}$ – $7.5 \times 10^{14}$ Hz
- Ultraviolet: $7.5 \times 10^{14}$ – $10^{16}$ Hz
- X-rays and beyond: $> 10^{16}$ Hz
Our frequency $2.66 \times 10^{15}$ Hz falls squarely in the ultraviolet range — well above visible light, but below X-rays.

Watch out

A common mistake is to forget the unit conversion and plug 11 eV directly into $E = h\nu$ with $h$ in J·s. That gives a wildly wrong answer. Always convert to joules first, or use $h = 4.1357 \times 10^{-15} \, \text{eV·s}$ if you prefer working in eV — then $\nu = 11 / (4.1357 \times 10^{-15}) \approx 2.66 \times 10^{15}$ Hz, same result.

Tip

For quick estimation, remember that 1 eV corresponds to a frequency of about $2.42 \times 10^{14}$ Hz (since $1 \, \text{eV} / h \approx 2.42 \times 10^{14}$ Hz). So 11 eV gives roughly $11 \times 2.42 \times 10^{14} = 2.66 \times 10^{15}$ Hz — no calculator needed for the order of magnitude.

✓Final answer

2.66 \times 10^{15}

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
$\lambda = \frac{1.989}{3.3} \times 10^{-6}$

Do the division

$\frac{1.989}{3.3} = 0.6027\ldots$
Rounding to two significant figures (since $3.3$ has two significant figures):
$\lambda \approx 0.60 \times 10^{-6}\ \text{m} = 6.0 \times 10^{-7}\ \text{m}$
Watch out

A common mistake is to forget that $E$ is in joules and $h$ is in J·s — if you accidentally use $h$ in eV·s, you’ll get a wildly wrong answer. Always check units: J·s × m/s = J·m, then dividing by J gives metres. Clean.

Convert to nanometres (optional but helpful for context) $1\ \text{nm} = 10^{-9}\ \text{m}$ , so

$\lambda = 6.0 \times 10^{-7}\ \text{m} = 600\ \text{nm}$
This wavelength lies in the visible spectrum — specifically, orange light.

✓Final answer
$6.0 \times 10^{-7}\ \text{m}$
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.

(Cross-check via wavelength: $\lambda = \dfrac{1240\ \text{eV·nm}}{13200\ \text{eV}} \approx 0.094$ nm ≈ 0.94 Å, a typical X-ray wavelength.)

✓Final answer
$(b) X-rays.$

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Q.One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

Concept understanding — Photon Energy Calculation

Photon Energy Calculation

The Core Idea

The Master Formula

Working in Electron-Volts

Worked Example 1 — from frequency

Worked Example 2 — from wavelength

Total Energy of a Beam

Summary

Why this formula?

Photon Energy Calculation

The Working Formula

Two Handy Shortcuts

Worked Idea

Method: Finding the Minimum Frequency to Supply a Given Photon Energy (and Identifying the Spectral Band)

Steps

Q.One requires 11 eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in

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