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.

AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$Q. To dissociate a molecule into its component atoms, the energy required is 13.2 eV. The frequency of the electromagnetic radiation corresponding to this energy lies in (A) infrared region (B) visible region (C) microwave region (D) ultraviolet region$
›Reveal solutionSolution
Converting the 13.2 eV dissociation energy to a photon frequency/wavelength via $E=hf=hc/\lambda$ gives a wavelength around 94 nm, which lies in the ultraviolet region.

Concept and Intuition

Each photon of electromagnetic radiation carries energy $E=hf=\dfrac{hc}{\lambda}$ . To identify which part of the spectrum corresponds to a given bond/dissociation energy, convert that energy into a wavelength and compare it against the known spectral ranges (radio, microwave, infrared, ~400–700 nm visible, ultraviolet below ~400 nm, X-rays, etc.).

Step-by-Step Solution

Convert energy to joules: $E=13.2\ \mathrm{eV}\times1.6\times10^{-19}\ \mathrm{J/eV}=2.112\times10^{-18}$ J.

Find frequency: $f=\dfrac{E}{h}=\dfrac{2.112\times10^{-18}}{6.63\times10^{-34}}\approx3.19\times10^{15}$ Hz.

Find wavelength: $\lambda=\dfrac{c}{f}=\dfrac{3\times10^8}{3.19\times10^{15}}\approx9.4\times10^{-8}$ m $=94$ nm.

Since visible light spans roughly 400–700 nm and infrared/microwave are much longer wavelengths, a wavelength of ~94 nm (shorter than 400 nm) falls in the ultraviolet region.

Common Mistakes

Forgetting to convert eV to joules before using $E=hf$ .

Misjudging the visible-to-UV boundary (visible light starts around 400 nm; anything shorter is UV).

✓Final answer
$The correct option is (D) — ultraviolet region.$

ANSWER: D
AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$5 \times 10^{14}$
›Reveal solutionSolution
Tests converting beam power into a photon flux using $E=hf$ ; the answer is $10\times10^{16}$ photons/s.

Concept and Intuition

A beam's power is the total energy it delivers per second. If each photon of frequency $f$ carries a fixed quantum of energy $E=hf$ , then the number of photons crossing per second is simply the power divided by the energy of one photon: $N = P/E$ .

Step-by-Step Solution

Energy of one photon: $E = hf = (6.6\times10^{-34})(5\times10^{14}) = 3.3\times10^{-19}$ J.

Convert power to SI: $P = 33\ \mathrm{mW} = 3.3\times10^{-2}$ W.

Photon rate: $N = \dfrac{P}{E} = \dfrac{3.3\times10^{-2}}{3.3\times10^{-19}} = 1\times10^{17}$ photons/s.

Express in the option's form: $1\times10^{17} = 10\times10^{16}$ .

Common Mistakes

Forgetting to convert mW to W before dividing.

Mis-tracking the powers of ten when dividing $10^{-2}$ by $10^{-19}$ .

✓Final answer
$10 \times 10^{16}$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$Q. The minimum wavelength of X-rays produced by 20 kV electrons is nearly (A) 0.62 Å (B) 1.8 Å (C) 3.2 Å (D) 6.5 Å$
›Reveal solutionSolution
This tests the Duane-Hunt law for the short-wavelength (minimum wavelength) limit of the continuous X-ray spectrum; for 20 kV accelerating voltage, $\lambda_{min}=0.62$ Å.

Concept and Intuition

When high-speed electrons strike a target, some lose their entire kinetic energy in a single collision, emitting one photon carrying all of it — this photon has the maximum possible energy and hence the minimum possible wavelength in the continuous (bremsstrahlung) X-ray spectrum. Since the electron's kinetic energy comes from being accelerated through potential difference $V$ , energy conservation gives $eV = \dfrac{hc}{\lambda_{min}}$ .

Step-by-Step Solution

From $eV = \dfrac{hc}{\lambda_{min}}$ : $\lambda_{min} = \dfrac{hc}{eV}$ .

Using the standard combination $\dfrac{hc}{e} \approx 12400\,\mathrm{eV\,\mathring A}$ (i.e. $hc/e$ expressed conveniently in eV·Å), $\lambda_{min} = \dfrac{12400}{V(\mathrm{in\ volts})}\,\mathrm{\mathring A}$ .

With $V = 20\,\mathrm{kV} = 20000\,\mathrm{V}$ : $\lambda_{min} = \dfrac{12400}{20000} = 0.62\,\mathrm{\mathring A}$ .

Common Mistakes

Forgetting to convert kV to V before dividing.

Confusing the minimum-wavelength (continuum, energy-conservation) limit with the characteristic line wavelengths, which instead depend on the target element via Moseley's law.

✓Final answer
$The correct option is (A) — 0.62 Å.$

ANSWER: A
AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$10^{37}$
›Reveal solutionSolution
Divide the transmitted power by the energy of a single photon at the given wavelength. Answer: order $10^{31}$ .

Concept and Intuition

A radio transmitter emits energy as a stream of photons, each carrying energy $E = hc/\lambda$ . The number of photons emitted per second is simply the total power divided by the energy per photon, since power is energy delivered per unit time.

Step-by-Step Solution

$P = 10\ \mathrm{kW} = 10^4\ \mathrm{W}$ , $\lambda = 500\ \mathrm{m}$ .

$E_{photon} = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{500} = 3.98\times10^{-28}\ \mathrm{J}$ .

$n = \dfrac{P}{E_{photon}} = \dfrac{10^4}{3.98\times10^{-28}} \approx 2.5\times10^{31}$ photons/second.

Order of magnitude: $10^{31}$ .

Common Mistakes

Forgetting radio photons have very low individual energy (long wavelength), so the photon count is astronomically large.

Arithmetic slip in exponent handling when dividing $10^4$ by a number of order $10^{-28}$ .

✓Final answer
$10^{31}$

ANSWER: B
AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$27.17\times10^{18}$
›Reveal solutionSolution
Only 8% of the rated power actually becomes light; dividing this useful power by the energy of one photon (from $E=hc/\lambda$ ) gives the photon emission rate.

Concept and Intuition

The lamp's rated 150 W is the total electrical power drawn, but only a fraction (the stated efficiency) is converted into visible light — the rest is lost as heat. To find how many photons are emitted per second, we need the light power, then divide by the energy carried by a single photon of the given wavelength.

Step-by-Step Solution

Useful (optical) power: $P_{light} = 0.08\times150 = 12$ W.

Photon energy: $E = \dfrac{hc}{\lambda} = \dfrac{6.63\times10^{-34}\times3\times10^8}{4500\times10^{-10}}$ .

Numerator: $6.63\times3=19.89 \Rightarrow 19.89\times10^{-26}$ . Denominator: $4.5\times10^{-7}$ .

$E = \dfrac{19.89\times10^{-26}}{4.5\times10^{-7}} = 4.42\times10^{-19}$ J.

Photons/sec $=\dfrac{P_{light}}{E}=\dfrac{12}{4.42\times10^{-19}}\approx2.717\times10^{19}=27.17\times10^{18}$ .

Common Mistakes

Using the full 150 W instead of the 8%-efficient 12 W.

Arithmetic slip in the powers of ten when dividing by $\lambda$ in angstroms — always convert to metres first ( $4500\,\text{Å}=4.5\times10^{-7}$ m).

✓Final answer
$27.17\times10^{18}$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$6.6\times10^{-3}$
›Reveal solutionSolution
This is a direct photon-counting problem: divide laser power by the energy of a single photon to get the photon emission rate, $2\times10^{16}$ per second.

Concept and Intuition

A laser's power output is just the total energy emitted per second, which equals (number of photons per second) × (energy per photon). Since each photon of wavelength $\lambda$ carries energy $E=hc/\lambda$ (or $h\nu$ ), the photon rate is simply power divided by this per-photon energy.

Step-by-Step Solution

Energy per photon: $E = \dfrac{hc}{\lambda} = \dfrac{(6.6\times10^{-34})(3\times10^8)}{6\times10^{-7}}$ .

Numerator: $6.6\times10^{-34}\times3\times10^8 = 19.8\times10^{-26} = 1.98\times10^{-25}$ J·m.

$E = \dfrac{1.98\times10^{-25}}{6\times10^{-7}} = 0.33\times10^{-18} = 3.3\times10^{-19}$ J.

Photon rate: $n = \dfrac{P}{E} = \dfrac{6.6\times10^{-3}}{3.3\times10^{-19}} = 2\times10^{16}$ photons per second.

Common Mistakes

Using $\lambda$ in nanometres without converting to metres.

Confusing power (energy per second) with energy per photon and dividing the wrong way round.

✓Final answer
$2\times10^{16}$

ANSWER: A
AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
$\nu$
›Reveal solutionSolution
Combining the photon energy $E=h\nu$ with the relativistic energy-momentum relation for a massless particle, $p=E/c$ , gives $p=h\nu/c$ .

Concept and Intuition

A photon has zero rest mass, so its total energy and momentum are related simply by $E=pc$ (the massless limit of $E^2=(pc)^2+(mc^2)^2$ ). Since a photon of frequency $\nu$ carries energy $E=h\nu$ (Planck's relation), substituting gives its momentum directly.

Step-by-Step Solution

Photon energy: $E=h\nu$ .

Massless-particle relation: $E=pc$ , so $p=\dfrac{E}{c}$ .

Substitute: $p=\dfrac{h\nu}{c}$ .

Common Mistakes

Writing $p=h\nu/c^2$ (confusing this with mass-energy equivalence $m=E/c^2$ , which is not the same as momentum).

Forgetting the factor of $c$ altogether and writing $p=h\nu$ .

✓Final answer
$\dfrac{h\nu}{c}$

ANSWER: D
AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
$1.50 \times 10^{25}$
›Reveal solutionSolution
This tests the conversion of a macroscopic radio-wave power output into a photon count, using $E_{photon} = hf$ and $N = P/E_{photon}$ . The answer is $1.72\times10^{31}$ photons per second.

Concept and Intuition

Even though radio waves behave classically at the level of everyday antennas, the electromagnetic energy they carry is still quantized into photons of energy $E = hf$ . Because radio-frequency photons are extremely low-energy (since $f$ is small), an ordinary transmitter emits an enormous number of them per second to deliver even modest total power — this is the conceptual bridge between the quantum picture (individual photons) and the classical picture (continuous power output).

Step-by-Step Solution

Frequency: $f = 880\ \text{kHz} = 8.8\times10^5\ \text{Hz}$ .

Energy per photon: $E = hf = (6.626\times10^{-34})(8.8\times10^5) = 5.831\times10^{-28}\ \text{J}$ .

Total power: $P = 10\ \text{kW} = 1\times10^4\ \text{W} = 1\times10^4\ \text{J/s}$ .

Number of photons emitted per second: $N = \dfrac{P}{E} = \dfrac{10^4}{5.831\times10^{-28}} \approx 1.715\times10^{31} \approx 1.72\times10^{31}$ .

Common Mistakes

Forgetting to convert kHz to Hz (using $880$ instead of $8.8\times10^5$ ), which throws off the answer by many orders of magnitude.

Mixing up $P\times E$ and $P/E$ — since we want the count of photons carrying a fixed total power, we must divide the power by the energy per photon, not multiply.

✓Final answer
$1.72 \times 10^{31}$

ANSWER: C
AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$Q. The shortest wavelength of X-rays emitted from an X-ray tube depends upon _______ (A) Nature of the gas in the tube (B) Voltage applied to tube (C) Current in the tube (D) Nature of target of the tube$
›Reveal solutionSolution
The cut-off wavelength of the continuous (bremsstrahlung) X-ray spectrum is fixed purely by the accelerating voltage across the tube, since it corresponds to an electron converting its entire kinetic energy into a single photon.

Concept and Intuition

In an X-ray tube, electrons are accelerated through a potential difference $V$ and slammed into a target. Most electrons undergo multiple deceleration events, radiating a continuous spectrum of photon energies (Bremsstrahlung), but the maximum possible photon energy — and hence the minimum possible wavelength — occurs when an electron loses all of its kinetic energy $eV$ in a single collision, emitting one photon of energy $eV = h\nu_{max} = \dfrac{hc}{\lambda_{min}}$ . This depends only on how much energy the electron was given by the accelerating voltage, not on which target it hits or how much current flows.

Step-by-Step Solution

Kinetic energy gained by an electron accelerated through voltage $V$ : $KE = eV$ .

Maximum photon energy possible from a single electron = its entire KE: $h\nu_{max}=eV$ .

$\lambda_{min}=\dfrac{hc}{eV}$ — depends only on $V$ (and universal constants $h,c,e$ ).

Target material affects the characteristic line spectrum (specific line wavelengths), and tube current affects intensity, but neither affects $\lambda_{min}$ .

Common Mistakes

Confusing the characteristic spectrum (depends on target material) with the continuous spectrum's cut-off wavelength (depends only on voltage).

Thinking higher tube current gives shorter $\lambda_{min}$ — current only changes intensity (number of photons), not their maximum energy.

✓Final answer
$The correct option is (B) — Voltage applied to tube.$

ANSWER: B

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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

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